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Cambridge IGCSE Mathematics 0580 · Core and Extended · Topic 3

Coordinate geometry

179 free, high-quality past-paper-style questions on Coordinate geometry for Cambridge IGCSE Maths, each with a full mark scheme. Covers: Gradient, midpoint and length of a line segment; Equation of a straight line, parallel and perpendicular lines. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,193 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

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+ Add to test✓ In your test1IGMC-043-00-01Gradient, midpoint and length of a line segment Paper CEasy4 marks

No calculator

The point A has coordinates (2, 5) and the point B has coordinates (8, 1).

Diagram not to scale

(a)

Find the coordinates of the midpoint of AB.

(2)

(b)

Find the gradient of the line AB. Give your answer as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{2+8}{2}\) or \(\frac{5+1}{2}\)	✓ 1
A1 (5, 3)	✓ 1
Part (b)
M1 for \(\frac{1-5}{8-2}\) or equivalent	✓ 1
A1 \(-\frac{2}{3}\)	✓ 1

Common error: Subtracting the coordinates in a different order on the top and bottom gives \(+\frac{2}{3}\). Take both differences from the same point.

+ Add to test✓ In your test2IGMC-043-00-02Gradient, midpoint and length of a line segment Paper CEasy5 marks

Calculator

The point P has coordinates (−1, 3) and the point Q has coordinates (6, 8).

Diagram not to scale

(a)

Find the coordinates of the midpoint of PQ.

(2)

(b)

Calculate the length of PQ. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{-1+6}{2}\) or \(\frac{3+8}{2}\)	✓ 1
A1 (2.5, 5.5)	✓ 1
Part (b)
M1 for horizontal difference 7 and vertical difference 5	✓ 1
M1 for \(\sqrt{7^2+5^2}\) or \(\sqrt{74}\)	✓ 1
A1 8.60	✓ 1

Exam tip: Write down \(\sqrt{74}\) before rounding. An answer of 8.6 is not to 3 significant figures; write 8.60.

+ Add to test✓ In your test3IGMC-043-00-03Equation of a straight line, parallel and perpendicular lines Paper CEasy5 marks

No calculator

A straight line passes through the points (0, −2) and (4, 10).

(a)

Find the gradient of the line.

(2)

(b)

Write down the equation of the line in the form y = mx + c.

(2)

(c)

Show that the point (5, 13) lies on the line.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{10-(-2)}{4-0}\)	✓ 1
A1 3	✓ 1
Part (b)
B1 for c = −2 or m = 3 used in an equation	✓ 1
B1 y = 3x − 2	✓ 1
Part (c)
B1 3 × 5 − 2 = 13 shown, with conclusion	✓ 1

Common error: Writing \(10-2\) instead of \(10-(-2)\) gives a gradient of 2. Put negative coordinates in brackets.

+ Add to test✓ In your test4IGMC-043-00-04Gradient, midpoint and length of a line segment Paper CEasy4 marks

No calculator

The point M(4, 1) is the midpoint of the line segment AB. The point A has coordinates (1, −3).

(a)

Find the coordinates of B.

(2)

(b)

Find the length of AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{1+x}{2}=4\) or \(\frac{-3+y}{2}=1\), or a step of (+3, +4) from A to M	✓ 1
A1 (7, 5)	✓ 1
Part (b)
M1 for \(\sqrt{6^2+8^2}\)	✓ 1
A1 10	✓ 1

Common error: Averaging A and M gives (2.5, −1), which is not B. M is halfway, so add the step from A to M again to reach B.

+ Add to test✓ In your test5IGMC-043-00-05Gradient, midpoint and length of a line segment Paper CEasy5 marks

Calculator

The point C has coordinates (1.5, 2.8) and the point D has coordinates (4.7, −3.6).

Diagram not to scale

(a)

Find the gradient of CD.

(2)

(b)

Calculate the length of CD. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{-3.6-2.8}{4.7-1.5}\)	✓ 1
A1 −2	✓ 1
Part (b)
M1 for 3.2 and 6.4 (or −6.4) as the differences	✓ 1
M1 for \(\sqrt{3.2^2+6.4^2}\) or \(\sqrt{51.2}\)	✓ 1
A1 7.16	✓ 1

Exam tip: A negative difference becomes positive when it is squared. Gradients keep their sign; lengths are always positive.

+ Add to test✓ In your test6IGMC-043-01-01Equation of a straight line, parallel and perpendicular lines Paper CEasy3 marks

No calculator

A straight line has equation \(y = 3x - 5\).

(a)

Write down the gradient of the line.

(1)

(b)

Write down the coordinates of the point where the line crosses the y-axis.

(1)

(c)

Show that the point (4, 7) lies on the line.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 3	✓ 1
Part (b)
B1: (0, −5)	✓ 1
Part (c)
B1: substitutes x = 4 to get 3 × 4 − 5 = 7, which equals the y-coordinate, so the point lies on the line	✓ 1

Exam tip: In \(y = mx + c\), the number multiplying x is the gradient and the constant term is the y-intercept. Give the intercept as coordinates when the question asks for a point.

+ Add to test✓ In your test7IGMC-043-01-02Equation of a straight line, parallel and perpendicular lines Paper CEasy4 marks

No calculator

The points A and B have coordinates A(2, 1) and B(6, 9).

(a)

Find the gradient of the line AB.

(2)

(b)

Find the equation of the line AB. Give your answer in the form \(y = mx + c\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(\frac{9-1}{6-2}\)	✓ 1
A1: 2	✓ 1
Part (b)
M1: substitutes a point into y = 2x + c, e.g. 1 = 2 × 2 + c	✓ 1
A1: y = 2x − 3	✓ 1

Common error: Keep the order of subtraction the same on the top and bottom of the gradient fraction. Mixing the order gives the wrong sign.

+ Add to test✓ In your test8IGMC-043-01-04Equation of a straight line, parallel and perpendicular lines Paper CEasy4 marks

No calculator

A line has equation \(2y + 6x = 8\).

(a)

Rearrange the equation into the form \(y = mx + c\).

(2)

(b)

Write down the gradient of the line.

(1)

(c)

Write down the coordinates of the point where the line crosses the y-axis.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 2y = −6x + 8	✓ 1
A1: y = −3x + 4	✓ 1
Part (b)
B1: −3	✓ 1
Part (c)
B1: (0, 4)	✓ 1

Common error: Do not read the gradient as 6 from the original equation. Make y the subject first, then read off the coefficient of x.

+ Add to test✓ In your test9IGMC3-3-09c-01Gradient, midpoint and length of a line segment Paper CEasy7 marks

Calculator

A survey drone flies in a straight line from take-off point A(−1, 2) to a landing point B(7, 7). Distances on the map are in kilometres.

(a)

Calculate the gradient of the line AB.

(2)

(b)

Find the coordinates of the midpoint of AB.

(2)

(c)

Calculate the length of AB.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 change in y divided by change in x, with 5 and 8 identified	✓ 1
A1 gradient = 5/8	✓ 1
Part (b)
B1 x-coordinate 3	✓ 1
B1 y-coordinate 4.5, so (3, 4.5)	✓ 1
Part (c)
M1 horizontal difference 8 and vertical difference 5	✓ 1
M1 √(8² + 5²) = √89	✓ 1
A1 9.43 km	✓ 1

Exam tip: Write the two differences before squaring; subtracting coordinates in a consistent order avoids sign slips.

+ Add to test✓ In your test10IGMC3-3-09c-02Gradient, midpoint and length of a line segment Paper CEasy6 marks

No calculator

During a hospital drip, the volume V ml of fluid delivered after t minutes is modelled by a straight line. The line passes through the points (2, 13) and (6, 25) on a graph of V against t.

(a)

Find the gradient of the line.

(2)

(b)

Find the equation of the line in the form V = mt + c.

(2)

(c)

Use your equation to find the time when 100 ml has been delivered.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (25 − 13) ÷ (6 − 2)	✓ 1
A1 gradient = 3	✓ 1
Part (b)
M1 substitutes a point into V = 3t + c, e.g. 13 = 6 + c	✓ 1
A1 V = 3t + 7	✓ 1
Part (c)
M1 100 = 3t + 7 (or 93 ÷ 3)	✓ 1
A1 t = 31 minutes	✓ 1

Common error: Reading the gradient as a ratio of x-change to y-change turns 3 into 1/3; always put the change in y on top.

+ Add to test✓ In your test11IGMC3-3-09c-05Gradient, midpoint and length of a line segment Paper CEasy5 marks

Calculator

A relay balloon hovers above the point M(1, −2), which is the midpoint of two ground stations P(−3, 4) and Q. Coordinates are in kilometres.

Diagram not to scale

(a)

Find the coordinates of Q.

(3)

(b)

Calculate the distance PQ.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses midpoint relation, e.g. (−3 + x)/2 = 1 or the step from P to M repeated	✓ 1
A1 x = 5	✓ 1
A1 y = −8, so Q = (5, −8)	✓ 1
Part (b)
M1 √(8² + 12²) = √208	✓ 1
A1 14.4 km	✓ 1

Exam tip: Check your answer for Q by finding the midpoint of P and Q again; it should return M.

+ Add to test✓ In your test12IGMC3-3-09c-09Equation of a straight line, parallel and perpendicular lines Paper CEasy6 marks

No calculator

On a map of a running route, a straight section of path lies along the line y = 3x − 7.

(a)

Write down the gradient and the y-intercept of the line.

(2)

(b)

Does the point (4, 5) lie on the path? Show how you decide.

(2)

(c)

A drinks station is on the path at the point where y = 11. Find its x-coordinate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient = 3	✓ 1
B1 y-intercept = −7	✓ 1
Part (b)
M1 substitutes x = 4 to get 3(4) − 7	✓ 1
A1 = 5, so yes, it lies on the path	✓ 1
Part (c)
M1 11 = 3x − 7	✓ 1
A1 x = 6	✓ 1

Common error: The y-intercept is the constant term including its sign, so here it is −7, not 7.

+ Add to test✓ In your test13IGMC3-3-09r5-01Gradient, midpoint and length of a line segment Paper CEasy6 marks

Calculator

A beekeeper marks two hives at P(1, 2) and Q(7, 10) on a plan. Each unit on the plan is 1 metre.

(a)

Calculate the gradient of the line PQ.

(2)

(b)

Find the coordinates of the midpoint of PQ.

(2)

(c)

Calculate the distance between the two hives.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 change in y divided by change in x, (10 − 2)/(7 − 1)	✓ 1
A1 4/3 (or 1.33 to 3 s.f.)	✓ 1
Part (b)
B1 x = 4	✓ 1
B1 y = 6	✓ 1
Part (c)
M1 √(6² + 8²)	✓ 1
A1 10 m	✓ 1

Exam tip: For a length, square both differences first, add, then take the square root; do not square root each difference.

+ Add to test✓ In your test14IGMC3-3-09r5-02Equation of a straight line, parallel and perpendicular lines Paper CEasy4 marks

Calculator

At a honey stall, a straight price-guide line passes through the points (0, −4) and (3, 5).

(a)

Find the gradient of the line.

(2)

(b)

Write down the equation of the line.

(1)

(c)

Find the value of y on the line when x = 7.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (5 − (−4))/(3 − 0)	✓ 1
A1 3	✓ 1
Part (b)
B1 y = 3x − 4	✓ 1
Part (c)
B1 y = 3 × 7 − 4 = 17	✓ 1

Common error: Reading the y-intercept from the wrong point: the intercept is the value of y where x = 0.

+ Add to test✓ In your test15IGMC3-3-09r5-03Gradient, midpoint and length of a line segment Paper CEasy6 marks

Calculator

On a hospital site plan, the ambulance base is at A(−3, 2) and the main entrance is at H(5, −4). One unit on the plan is 1 km.

Diagram not to scale

(a)

Find the midpoint of AH.

(2)

(b)

Calculate the straight-line distance AH.

(2)

(c)

An ambulance travels along AH at a constant 60 km/h. Calculate the time taken, in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x = 1	✓ 1
B1 y = −1	✓ 1
Part (b)
M1 √(8² + 6²)	✓ 1
A1 10 km	✓ 1
Part (c)
M1 time = 10 ÷ 60 hours (ECF from (b))	✓ 1
A1 10 minutes	✓ 1

Exam tip: Convert hours to minutes at the end by multiplying by 60.

+ Add to test✓ In your test16IGMC3-3-09r5-04Equation of a straight line, parallel and perpendicular lines Paper CEasy4 marks

No calculator

A barrier in a Formula 1 pit lane is drawn on a plan along the line y = −2x + 9.

(a)

Write down the gradient and the y-intercept of the line.

(2)

(b)

Find the x-coordinate of the point where the line crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient = −2	✓ 1
B1 y-intercept = 9	✓ 1
Part (b)
M1 substitute y = 0, so 2x = 9	✓ 1
A1 x = 4.5	✓ 1

Exam tip: On the x-axis the y-coordinate is zero; on the y-axis the x-coordinate is zero.

+ Add to test✓ In your test17IGMC3-3-09r8-01Gradient, midpoint and length of a line segment Paper CEasy6 marks

No calculator

Two research buoys, A and B, are anchored in a bay. On a map grid with units of 1 km, A is at (2, 3) and B is at (8, 11).

(a)

Find the gradient of the line AB.

(2)

(b)

A third buoy, M, is placed halfway between A and B. Find the coordinates of M.

(2)

(c)

Calculate the distance AB, in km.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at (11 − 3)/(8 − 2)	✓ 1
A1 gradient = 4/3	✓ 1
Part (b)
M1 attempt at ((2 + 8)/2, (3 + 11)/2)	✓ 1
A1 M = (5, 7)	✓ 1
Part (c)
M1 √((8 − 2)² + (11 − 3)²) = √(36 + 64)	✓ 1
A1 10 km	✓ 1

Exam tip: Write the substitution into the distance formula before evaluating; method marks depend on seeing it.

+ Add to test✓ In your test18IGMC3-3-09r8-02Equation of a straight line, parallel and perpendicular lines Paper CEasy6 marks

No calculator

In an architect's plan, a roof beam is a straight line with gradient −2 that passes through the point (0, 5). All units are metres.

(a)

Write down the equation of the beam in the form y = mx + c.

(2)

(b)

Show that the point (4, −3) lies on the beam.

(2)

(c)

Find the x-coordinate of the point where the beam meets the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 m = −2 used	✓ 1
B1 y = −2x + 5	✓ 1
Part (b)
M1 substitute x = 4: −2(4) + 5	✓ 1
A1 = −3, so the point lies on the beam	✓ 1
Part (c)
M1 0 = −2x + 5	✓ 1
A1 x = 2.5 m	✓ 1

Common error: Setting x = 0 instead of y = 0 when asked for the x-axis crossing.

+ Add to test✓ In your test19IGMC3-3-09r8-03Equation of a straight line, parallel and perpendicular lines Paper CEasy6 marks

No calculator

A delivery drone flies in a straight line over a map grid. It passes over P(1, 4) and then over Q(5, 12).

(a)

Find the gradient of PQ.

(2)

(b)

Find the equation of the line PQ.

(2)

(c)

The drone continues along the same line. Find its y-coordinate when x = 8.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (12 − 4)/(5 − 1)	✓ 1
A1 2	✓ 1
Part (b)
M1 y − 4 = 2(x − 1) or substitutes a point into y = 2x + c	✓ 1
A1 y = 2x + 2	✓ 1
Part (c)
M1 substitute x = 8 into their equation	✓ 1
A1 y = 18	✓ 1

Exam tip: Check your equation by substituting the other given point.

+ Add to test✓ In your test20IGMC3-3-09r8-04Gradient, midpoint and length of a line segment Paper CEasy4 marks

No calculator

Two tide gauges are at A(−3, 2) and B(7, −6) on a coastal survey grid. M is the midpoint of AB.

Diagram not to scale

(a)

Find the coordinates of M.

(2)

(b)

A third gauge is at C(−1, 3). M is also the midpoint of CD. Find the coordinates of D.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ((−3 + 7)/2, (2 + (−6))/2)	✓ 1
A1 M = (2, −2)	✓ 1
Part (b)
M1 uses 2 × 2 − (−1) or 2 × (−2) − 3	✓ 1
A1 D = (5, −7)	✓ 1

Common error: Halving instead of doubling when working backwards from a midpoint to an end point.

+ Add to test✓ In your test21IGMC3-3-09r12-01Gradient, midpoint and length of a line segment Paper CEasy4 marks

No calculator

Two marker buoys in a marine reserve are placed at A(2, 1) and B(8, 13) on a coordinate grid. Each unit on the grid represents 1 km.

Diagram not to scale

(a)

Find the coordinates of the midpoint of AB.

(2)

(b)

Find the gradient of the line AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use \(\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\)	✓ 1
A1 (5, 7)	✓ 1
Part (b)
M1 \(\frac{13-1}{8-2}\) or equivalent	✓ 1
A1 2	✓ 1

Exam tip: Write the subtraction in the same order on the top and bottom of the gradient fraction.

+ Add to test✓ In your test22IGMC3-3-09r12-02Gradient, midpoint and length of a line segment Paper CEasy5 marks

Calculator

A tagged whale surfaces first at P(−3, 2) and later at Q(6, 14). Each unit on the grid represents 1 km.

(a)

Calculate the distance PQ.

(3)

(b)

Assume the whale swam in a straight line from P to Q at a constant speed of 5 km/h. Calculate the time taken.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differences in coordinates 9 and 12	✓ 1
M1 \(\sqrt{9^2+12^2}\)	✓ 1
A1 15 km	✓ 1
Part (b)
M1 time = distance ÷ speed = 15 ÷ 5	✓ 1
A1 3 hours	✓ 1

Common error: When a coordinate is negative, subtracting it needs care: 6 − (−3) = 9, not 3.

+ Add to test✓ In your test23IGMC3-3-09r12-03Equation of a straight line, parallel and perpendicular lines Paper CEasy6 marks

No calculator

In a plan for a pavilion, a roof beam lies along the straight line through the points (2, 5) and (6, 13). Each unit represents 1 m and the ground is the line y = 0.

(a)

Find the gradient of the beam.

(2)

(b)

Find the equation of the line, giving your answer in the form y = mx + c.

(2)

(c)

If the beam were extended, it would meet the ground at a point. Find the x-coordinate of this point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{13-5}{6-2}\)	✓ 1
A1 2	✓ 1
Part (b)
M1 substitute a point and gradient 2 into y = mx + c, e.g. 5 = 2(2) + c	✓ 1
A1 y = 2x + 1	✓ 1
Part (c)
M1 substitute y = 0: 0 = 2x + 1	✓ 1
A1 x = −0.5 (or −1/2)	✓ 1

Exam tip: Check your equation by substituting the other given point; it should satisfy the equation too.

+ Add to test✓ In your test24IGMC3-3-09r12-04Gradient, midpoint and length of a line segment Paper CEasy5 marks

Calculator

A logistics company has a warehouse at W(4, 9) and a depot at D(−2, −1). Each unit on the grid represents 1 km. A transfer hub H is to be built at the midpoint of WD.

Diagram not to scale

(a)

Find the coordinates of H.

(2)

(b)

Calculate the distance from W to H, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\left(\frac{4+(-2)}{2},\frac{9+(-1)}{2}\right)\)	✓ 1
A1 (1, 4)	✓ 1
Part (b)
M1 differences in coordinates 3 and 5	✓ 1
M1 \(\sqrt{3^2+5^2}\)	✓ 1
A1 5.83 km	✓ 1

Common error: The midpoint is found by adding and halving; the distance is found with Pythagoras. Do not mix up the two methods.

+ Add to test✓ In your test25IGMC3-3-10r15-01Gradient, midpoint and length of a line segment Paper CEasy6 marks

No calculator

On a game map, a player starts at A(−1, 2) and a treasure chest is at B(5, 10). Distances are measured in map squares.

(a)

Find the coordinates of the midpoint of AB.

(2)

(b)

Find the gradient of the line AB.

(2)

(c)

Find the length of AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ((−1 + 5)/2, (2 + 10)/2)	✓ 1
A1 (2, 6)	✓ 1
Part (b)
M1 (10 − 2)/(5 − (−1))	✓ 1
A1 4/3	✓ 1
Part (c)
M1 (5 − (−1))² + (10 − 2)² = 36 + 64	✓ 1
A1 √100 = 10 map squares	✓ 1

Common error: Subtracting a negative coordinate needs care: 5 − (−1) = 6, not 4.

+ Add to test✓ In your test26IGMC3-3-10r15-02Equation of a straight line, parallel and perpendicular lines Paper CEasy6 marks

No calculator

A water pipe runs in a straight line through the points A(0, 5) and B(4, 17), where coordinates are in kilometres.

(a)

Find the gradient of the pipe.

(2)

(b)

Write down the equation of the line AB in the form y = mx + c.

(2)

(c)

A valve is at C(7, 26). Show that C lies on the pipe.

(1)

(d)

Another valve is on the pipe where x = 10. Find its y-coordinate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (17 − 5)/(4 − 0)	✓ 1
A1 3	✓ 1
Part (b)
B1 m = 3 (ECF from (a))	✓ 1
B1 c = 5, so y = 3x + 5	✓ 1
Part (c)
B1 3 × 7 + 5 = 26, so C is on the line	✓ 1
Part (d)
A1 y = 3 × 10 + 5 = 35	✓ 1

Exam tip: The point where the line crosses the y-axis, (0, 5) here, gives the value of c directly.

+ Add to test✓ In your test27IGMC3-3-10r15-03Gradient, midpoint and length of a line segment Paper CEasy5 marks

Calculator

A pharmacist records the concentration C mg/l of a drug in a patient's blood at t hours after a dose. The readings are (t, C) = (1, 8.4) and (5, 3.6). Assume the two points lie on a straight line.

(a)

Find the coordinates of the midpoint of the two readings.

(2)

(b)

Calculate the gradient of the line through the two readings.

(2)

(c)

Explain what the gradient in (b) tells you about the concentration of the drug.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ((1 + 5)/2, (8.4 + 3.6)/2)	✓ 1
A1 (3, 6)	✓ 1
Part (b)
M1 (3.6 − 8.4)/(5 − 1)	✓ 1
A1 −1.2	✓ 1
Part (c)
B1 the concentration decreases by 1.2 mg/l for each hour that passes	✓ 1

Exam tip: When interpreting a gradient, state the change in the vertical quantity for each one-unit increase in the horizontal quantity, with units.

+ Add to test✓ In your test28IGMC3-3-10r15-04Equation of a straight line, parallel and perpendicular lines Paper CEasy5 marks

No calculator

In a game, a laser beam travels along the line y = 2x − 3. A wall W is a straight line parallel to the beam and passes through the point (4, 1).

(a)

Write down the gradient of W.

(1)

(b)

Find the equation of W, giving your answer in the form y = mx + c.

(2)

(c)

Find the x-coordinate of the point where W meets the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2	✓ 1
Part (b)
M1 1 = 2 × 4 + c, or y − 1 = 2(x − 4)	✓ 1
A1 y = 2x − 7	✓ 1
Part (c)
M1 0 = 2x − 7	✓ 1
A1 x = 3.5	✓ 1

Exam tip: Parallel lines have equal gradients, so only the intercept c has to be found for the new line.

+ Add to test✓ In your test29IGMC3-3-10r21-01Gradient, midpoint and length of a line segment Paper CEasy6 marks

No calculator

On a map of a mountain range, with distances in kilometres, two huts are at A(1, 2) and B(7, 10).

(a)

Find the gradient of the line AB.

(2)

(b)

Find the coordinates of the midpoint of AB.

(2)

(c)

Find the length of AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{10-2}{7-1}\)	✓ 1
A1 \(\frac{4}{3}\)	✓ 1
Part (b)
M1 \(\left(\frac{1+7}{2},\frac{2+10}{2}\right)\)	✓ 1
A1 (4, 6)	✓ 1
Part (c)
M1 \(\sqrt{(7-1)^2+(10-2)^2}\)	✓ 1
A1 10 km	✓ 1

Exam tip: Subtract the coordinates in the same order in the numerator and denominator when finding a gradient, or the sign will be wrong.

+ Add to test✓ In your test30IGMC3-3-10r21-02Gradient, midpoint and length of a line segment Paper CEasy5 marks

No calculator

In a food science experiment a sample of soup is heated. Its temperature T °C is recorded after t minutes. The graph of T against t is a straight line through the points (5, 20) and (25, 60).

(a)

Find the gradient of the line.

(2)

(b)

Find the equation of the line in the form \(T=mt+c\).

(2)

(c)

Find the time at which the temperature is 70 °C.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{60-20}{25-5}\)	✓ 1
A1 2	✓ 1
Part (b)
M1 \(20=2\times5+c\) (or equivalent)	✓ 1
A1 \(T=2t+10\)	✓ 1
Part (c)
A1 30 minutes	✓ 1

Common error: Writing the gradient as 'change in x over change in y'. The gradient is the change in the vertical quantity divided by the change in the horizontal quantity.

+ Add to test✓ In your test31IGMC3-3-10r21-03Gradient, midpoint and length of a line segment Paper CEasy4 marks

Calculator

Two telecoms masts A and B are shown on a map grid with distances in kilometres. Mast A is at (−4, 7). A relay station M at (2, 3) is exactly halfway between A and B.

(a)

Find the coordinates of B.

(2)

(b)

Calculate the distance AB, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{-4+x}{2}=2\) and \(\frac{7+y}{2}=3\)	✓ 1
A1 (8, −1)	✓ 1
Part (b)
M1 \(\sqrt{(8-(-4))^2+(-1-7)^2}\)	✓ 1
A1 14.4 km	✓ 1

Exam tip: Check your answer by finding the midpoint of your two endpoints: it should return the given midpoint.

+ Add to test✓ In your test32IGMC3-3-10r21-04Equation of a straight line, parallel and perpendicular lines Paper CEasy5 marks

No calculator

A straight path on a hillside is modelled by the line \(y=0.4x+3\), where x and y are in kilometres.

(a)

Write down the gradient of the path and the y-coordinate where it crosses the y-axis.

(2)

(b)

The point (10, k) lies on the path. Find k.

(1)

(c)

Determine whether the point (35, 16) lies on the path. Show your working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient 0.4	✓ 1
B1 3	✓ 1
Part (b)
A1 \(k=0.4\times10+3=7\)	✓ 1
Part (c)
M1 \(0.4\times35+3=17\)	✓ 1
A1 17 ≠ 16, so (35, 16) is not on the path	✓ 1

Exam tip: To test whether a point lies on a line, substitute its x-coordinate and compare the result with its y-coordinate.

+ Add to test✓ In your test33IGMC3-3-11r29-01Gradient, midpoint and length of a line segment Paper CEasy6 marks

Calculator

On a map grid, 1 unit represents 1 km. A pharmacy at A(2, 5) and a clinic at B(8, 13) are linked by a straight road.

(a)

Calculate the gradient of the line AB.

(2)

(b)

Find the coordinates of the midpoint of AB.

(2)

(c)

Calculate the length of the road AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at (13 − 5)/(8 − 2)	✓ 1
A1 gradient = 4/3 (accept 1.33)	✓ 1
Part (b)
M1 attempt to average the x-coordinates and the y-coordinates	✓ 1
A1 (5, 9)	✓ 1
Part (c)
M1 √(6² + 8²)	✓ 1
A1 10 km	✓ 1

Common error: Subtract the coordinates in the same order for both x and y; mixing the order changes the sign of the gradient.

+ Add to test✓ In your test34IGMC3-3-11r29-02Gradient, midpoint and length of a line segment Paper CEasy6 marks

Calculator

In a water-supply network, 1 unit on the plan represents 1 km. A pumping station is at P(−2, 5) and a reservoir is at R. The junction J(4, 1) is the midpoint of PR.

(a)

Find the coordinates of R.

(2)

(b)

Find the gradient of PR.

(2)

(c)

Calculate the length of the pipe PR, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of midpoint: (−2 + x)/2 = 4 and (5 + y)/2 = 1	✓ 1
A1 R = (10, −3)	✓ 1
Part (b)
M1 (−3 − 5)/(10 − (−2))	✓ 1
A1 −2/3 (accept −0.667)	✓ 1
Part (c)
M1 √(12² + 8²) (or √208)	✓ 1
A1 14.4 km	✓ 1

Exam tip: Write down the substituted values before using the distance formula so a slip in a sign can still earn the method mark.

+ Add to test✓ In your test35IGMC3-3-11r29-03Equation of a straight line, parallel and perpendicular lines Paper CEasy5 marks

No calculator

In a computer game, a laser beam travels along a straight line through the points (1, 4) and (3, 10).

(a)

Find the gradient of the beam.

(2)

(b)

Find the equation of the line in the form y = mx + c.

(2)

(c)

The beam passes through the point (5, k). Find k.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (10 − 4)/(3 − 1)	✓ 1
A1 3	✓ 1
Part (b)
M1 y − 4 = 3(x − 1) or 4 = 3 × 1 + c	✓ 1
A1 y = 3x + 1	✓ 1
Part (c)
A1 k = 16 (ECF from (b))	✓ 1

Exam tip: Check your equation by substituting the second given point as well as the first.

+ Add to test✓ In your test36IGMC3-3-11r29-04Equation of a straight line, parallel and perpendicular lines Paper CEasy6 marks

Calculator

A pharmacist prepares a mixture using x millilitres of solution A and y millilitres of solution B. The amounts must satisfy 3x + 4y = 24. This is the equation of a straight line L.

(a)

Rearrange the equation into the form y = mx + c and write down the gradient of L.

(2)

(b)

Find the coordinates of the points where L crosses the x-axis and the y-axis.

(2)

(c)

The pharmacist uses 4 ml of solution A. Calculate the volume of solution B needed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4y = −3x + 24	✓ 1
A1 y = −(3/4)x + 6, gradient −3/4	✓ 1
Part (b)
B1 (8, 0)	✓ 1
B1 (0, 6)	✓ 1
Part (c)
M1 3(4) + 4y = 24	✓ 1
A1 3 ml	✓ 1

Common error: When rearranging, divide every term by the coefficient of y, including the constant on the right-hand side.

+ Add to test✓ In your test37IGMC3-3-11r35-01Gradient, midpoint and length of a line segment Paper CEasy6 marks

No calculator

A small airfield is drawn on a square grid in which 1 unit represents 1 km. A runway runs in a straight line from A(2, 3) to B(10, 9).

(a)

Find the coordinates of the midpoint of AB.

(2)

(b)

Find the gradient of AB.

(2)

(c)

Find the length of the runway.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x-coordinate 6	✓ 1
B1 y-coordinate 6, so midpoint (6, 6)	✓ 1
Part (b)
M1 (9 − 3) ÷ (10 − 2)	✓ 1
A1 3/4 (or 0.75)	✓ 1
Part (c)
M1 √(8² + 6²)	✓ 1
A1 10 km	✓ 1

Common error: When finding a gradient, subtract the coordinates in the same order on the top and the bottom.

+ Add to test✓ In your test38IGMC3-3-11r35-02Gradient, midpoint and length of a line segment Paper CEasy4 marks

No calculator

A survey drone flies in a straight line from S(7, −2) to T. The midpoint of ST is M(4, 1).

(a)

Find the coordinates of T.

(2)

(b)

Find the gradient of ST.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses a midpoint relationship, e.g. (7 + x) ÷ 2 = 4 or (−2 + y) ÷ 2 = 1	✓ 1
A1 T = (1, 4)	✓ 1
Part (b)
M1 (4 − (−2)) ÷ (1 − 7), or uses M and S (ECF from (a) if T is used)	✓ 1
A1 −1	✓ 1

Exam tip: Work with the midpoint as an average: the midpoint is halfway, so the end point is as far beyond M as S is before it.

+ Add to test✓ In your test39IGMC3-3-11r35-03Equation of a straight line, parallel and perpendicular lines Paper CEasy5 marks

Calculator

A music teacher charges a fixed registration fee plus the same price for every lesson. The total cost, £C, for n lessons lies on a straight line. Two lessons cost £45 and six lessons cost £85.

(a)

Find the gradient of the straight line through (2, 45) and (6, 85).

(2)

(b)

Write down an equation for C in terms of n.

(2)

(c)

Find the cost of nine lessons.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (85 − 45) ÷ (6 − 2)	✓ 1
A1 10	✓ 1
Part (b)
M1 C = 10n + c with a correct substitution, e.g. 45 = 20 + c (ECF from (a))	✓ 1
A1 C = 10n + 25	✓ 1
Part (c)
B1 £115 (ECF from (b))	✓ 1

Common error: The gradient is the cost of one lesson, and the intercept is the fixed fee. Do not swap them.

+ Add to test✓ In your test40IGMC3-3-11r35-04Equation of a straight line, parallel and perpendicular lines Paper CEasy5 marks

No calculator

An irrigation pipe lies along the line with equation 2x + y = 10, where x and y are distances in metres on a plan of a field.

(a)

Rearrange the equation into the form y = mx + c and write down the gradient of the pipe.

(2)

(b)

Write down the y-intercept of the line.

(1)

(c)

Find the coordinates of the point where the pipe crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 y = 10 − 2x	✓ 1
A1 gradient −2	✓ 1
Part (b)
B1 10 (ECF from (a))	✓ 1
Part (c)
M1 substitutes y = 0, e.g. 2x = 10	✓ 1
A1 (5, 0)	✓ 1

Exam tip: To find where a line crosses the x-axis, put y = 0; to find where it crosses the y-axis, put x = 0.

+ Add to test✓ In your test41IGMC3-3-10r43-01Gradient, midpoint and length of a line segment Paper CEasy6 marks

Calculator

A warehouse robot travels in a straight line from A(1, 2) to B(6, 9) on a grid where one unit represents 1 m.

(a)

Find the gradient of AB.

(2)

(b)

Find the coordinates of the midpoint of AB.

(2)

(c)

Calculate the distance AB. Give your answer in metres, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (9 − 2) / (6 − 1)	✓ 1
A1 7/5 (or 1.4)	✓ 1
Part (b)
B1 x = 3.5	✓ 1
B1 y = 5.5	✓ 1
Part (c)
M1 √(5² + 7²)	✓ 1
A1 8.60 m (√74 = 8.6023…)	✓ 1

Common error: Subtract the coordinates in the same order for both the x and y differences when finding a gradient.

+ Add to test✓ In your test42IGMC3-3-10r43-02Gradient, midpoint and length of a line segment Paper CEasy5 marks

No calculator

A research buoy is at A(−2, 4). A second buoy is at B. The harbour mooring M(3, 1) is the midpoint of AB. Coordinates are in kilometres.

(a)

Find the coordinates of B.

(2)

(b)

Find the length of AB. Give your answer in the form a√b, where a and b are integers and b is as small as possible.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (x + (−2)) / 2 = 3 and (y + 4) / 2 = 1	✓ 1
A1 B(8, −2)	✓ 1
Part (b)
M1 √((8 − (−2))² + (−2 − 4)²)	✓ 1
A1 √136	✓ 1
A1 2√34 km (≈ 11.66 km)	✓ 1

Exam tip: To find an end point from a midpoint, double the midpoint coordinates and subtract the known end point.

+ Add to test✓ In your test43IGMC3-3-10r43-08Gradient, midpoint and length of a line segment Paper CEasy6 marks

Calculator

A trawler leaves port at A(0, 0) and sails in straight lines to B(3, 4), then to C(3, 10), then to D(11, 4). Coordinates are in kilometres.

Diagram not to scale

(a)

Calculate the distance AB.

(2)

(b)

Find the distance BC.

(1)

(c)

Calculate the distance CD.

(2)

(d)

Find the total distance sailed from A to D.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(3² + 4²)	✓ 1
A1 5 km	✓ 1
Part (b)
B1 6 km (B and C have the same x-coordinate, so 10 − 4)	✓ 1
Part (c)
M1 √(8² + (−6)²)	✓ 1
A1 10 km	✓ 1
Part (d)
B1 21 km (5 + 6 + 10; ECF)	✓ 1

Exam tip: If two points share an x-coordinate, the line between them is vertical and the length is the difference in y.

+ Add to test✓ In your test44IGMC3-3-10r43-10Gradient, midpoint and length of a line segment Paper CEasy5 marks

Calculator

On a fashion pattern grid, A is the point (2, −1) and B is the point (k, 8). The gradient of AB is 3. One unit on the grid is 1 cm.

(a)

Find the value of k.

(3)

(b)

Calculate the length of AB. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (8 − (−1)) / (k − 2) = 3	✓ 1
M1 9 = 3(k − 2)	✓ 1
A1 k = 5	✓ 1
Part (b)
M1 √((5 − 2)² + (8 − (−1))²) (ECF from (a))	✓ 1
A1 9.49 cm	✓ 1

Common error: Take care with the signs when subtracting a negative coordinate: 8 − (−1) = 9.

+ Add to test✓ In your test45IGMC3-3-10r43-13Equation of a straight line, parallel and perpendicular lines Paper CEasy5 marks

Calculator

The depth of water, d metres, at a distance x km from the shore on a fishing ground is modelled by the straight line d = 4 + 1.5x.

(a)

Write down the gradient of the line and say what it represents.

(2)

(b)

Find the depth of the water 6 km from the shore.

(1)

(c)

Find the distance from the shore at which the depth is 25 m.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient = 1.5	✓ 1
B1 the depth increases by 1.5 m for each extra kilometre from the shore	✓ 1
Part (b)
B1 13 m	✓ 1
Part (c)
M1 25 = 4 + 1.5x	✓ 1
A1 14 km	✓ 1

Exam tip: Substitute the given value into the correct variable, then solve; check that the answer is sensible in the context.

+ Add to test✓ In your test46IGMC3-3-11r44-01Gradient, midpoint and length of a line segment Paper CEasy5 marks

Calculator

A health authority marks two clinics on a map grid where 1 unit represents 1 km. Clinic A is at (1, 2) and clinic B is at (8, 6).

(a)

A mobile testing laboratory is to be placed halfway between the clinics. Find its coordinates.

(2)

(b)

Calculate the distance between the two clinics. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses (1 + 8)/2 and (2 + 6)/2	✓ 1
A1 (4.5, 4)	✓ 1
Part (b)
M1 correct differences 7 and 4	✓ 1
M1 attempt at √(7² + 4²)	✓ 1
A1 8.06 km	✓ 1

Common error: When finding a midpoint, add the coordinates and divide by 2. Do not subtract them, which is the method for length and gradient.

+ Add to test✓ In your test47IGMC3-3-11r44-02Equation of a straight line, parallel and perpendicular lines Paper CEasy6 marks

No calculator

On a map of an offshore wind farm, 1 unit represents 1 km and the x-axis is the coastline. A cable runs in a straight line from P(1, 5) to Q(4, 11).

(a)

Find the gradient of the cable.

(2)

(b)

Find the equation of the line PQ, giving your answer in the form y = mx + c.

(2)

(c)

The line PQ is extended backwards until it meets the coastline. Find the x-coordinate of this point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (11 − 5)/(4 − 1)	✓ 1
A1 2	✓ 1
Part (b)
M1 substitutes a point and gradient 2 into y = mx + c	✓ 1
A1 y = 2x + 3	✓ 1
Part (c)
M1 sets y = 0 in 2x + 3 = 0	✓ 1
A1 x = −1.5 (accept −3/2)	✓ 1

Exam tip: To find where a line crosses the x-axis, set y = 0. To find where it crosses the y-axis, set x = 0.

+ Add to test✓ In your test48IGMC3-3-11r44-03Gradient, midpoint and length of a line segment Paper CEasy4 marks

Calculator

On a map of a retailer's delivery network, 1 unit represents 1 km. A depot is at A(−2, 5). A distribution hub at H(3, 1) is exactly halfway along the straight route from A to a second depot B.

(a)

Find the coordinates of B.

(2)

(b)

Calculate the length of the route AB. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses midpoint formula, e.g. (−2 + x)/2 = 3 and (5 + y)/2 = 1	✓ 1
A1 B(8, −3)	✓ 1
Part (b)
M1 √(10² + 8²) or √164	✓ 1
A1 12.8 km	✓ 1

Common error: The midpoint is not the other end of the line. To find a missing endpoint, double the midpoint coordinate and subtract the known endpoint coordinate.

+ Add to test✓ In your test49IGMC3-3-11r44-04Equation of a straight line, parallel and perpendicular lines Paper CEasy7 marks

No calculator

A model for an outbreak says that the number of new cases per day, y, falls steadily in a straight-line pattern. On day x = 2 there are 40 new cases and on day x = 8 there are 10 new cases.

(a)

Find the gradient of the line through (2, 40) and (8, 10).

(2)

(b)

State what the gradient means in this situation.

(1)

(c)

Find the equation of the line in the form y = mx + c.

(2)

(d)

Use the model to find the day on which the number of new cases reaches 0.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (10 − 40)/(8 − 2)	✓ 1
A1 −5	✓ 1
Part (b)
B1 new cases fall by 5 each day	✓ 1
Part (c)
M1 substitutes (2, 40) and gradient −5, e.g. 40 = −5(2) + c	✓ 1
A1 y = −5x + 50	✓ 1
Part (d)
M1 sets −5x + 50 = 0	✓ 1
A1 day 10	✓ 1

Exam tip: When a gradient is given a meaning, state the change in y for each 1 unit increase in x, with units.

+ Add to test✓ In your test50IGMC3-3-10r47-01Gradient, midpoint and length of a line segment Paper CEasy6 marks

Calculator

On a map of a mountain area, with units in kilometres, hut A is at (2, 3) and hut B is at (8, 11).

(a)

Find the gradient of the line AB. Give your answer as a fraction in its simplest form.

(2)

(b)

Find the coordinates of the midpoint of AB.

(2)

(c)

Calculate the distance between the huts.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (11 − 3) / (8 − 2)	✓ 1
A1 4/3	✓ 1
Part (b)
M1 ((2 + 8)/2, (3 + 11)/2)	✓ 1
A1 (5, 7)	✓ 1
Part (c)
M1 distance = √((8 − 2)² + (11 − 3)²)	✓ 1
A1 10 km	✓ 1

Exam tip: Subtract the coordinates in the same order for the x and y differences; the signs then cancel when squared.

+ Add to test✓ In your test51IGMC3-3-10r47-02Gradient, midpoint and length of a line segment Paper CEasy4 marks

No calculator

Radio tower A is at (−4, 3). The point M(2, 5.5) is the midpoint of the line joining A to a second tower B. All units are in kilometres.

(a)

Find the coordinates of B.

(2)

(b)

Work out the distance AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for example (−4 + x)/2 = 2 and (3 + y)/2 = 5.5, or an increase of 6 and 2.5 from A to M applied again	✓ 1
A1 B = (8, 8)	✓ 1
Part (b)
M1 distance = √(12² + 5²) (ECF from (a))	✓ 1
A1 13 km	✓ 1

Common error: The midpoint is halfway between the points, so B is as far beyond M as A is before it; do not copy the midpoint coordinates as B.

+ Add to test✓ In your test52IGMC3-3-10r47-03Gradient, midpoint and length of a line segment Paper CEasy5 marks

Calculator

A climber's height h metres above sea level is modelled by a straight line against time t minutes. The model passes through (10, 1200) and (40, 1650).

(a)

Find the gradient of the line and state what it represents.

(2)

(b)

Find the equation of the line in the form h = mt + c.

(2)

(c)

Use the model to find the height at t = 55.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (1650 − 1200) / (40 − 10)	✓ 1
A1 15, the climber gains 15 metres of height every minute	✓ 1
Part (b)
M1 1200 = 15 × 10 + c or h − 1200 = 15(t − 10)	✓ 1
A1 h = 15t + 1050	✓ 1
Part (c)
A1 1875 m (ECF from (b))	✓ 1

Exam tip: Check your equation by substituting the second point; it must give the other height.

+ Add to test✓ In your test53IGMC3-3-10r47-04Equation of a straight line, parallel and perpendicular lines Paper CEasy5 marks

Calculator

Telecoms cable L₁ follows the line y = 2x − 5. A second cable L₂ is parallel to L₁ and passes through the point (3, 5). Units are kilometres.

(a)

Write down the gradient of L₂.

(1)

(b)

Find the equation of L₂. Give your answer in the form y = mx + c.

(2)

(c)

Find the coordinates of the point where L₂ crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2	✓ 1
Part (b)
M1 y − 5 = 2(x − 3) or 5 = 2 × 3 + c	✓ 1
A1 y = 2x − 1	✓ 1
Part (c)
M1 substitute y = 0 to get 0 = 2x − 1	✓ 1
A1 (0.5, 0)	✓ 1

Common error: Parallel lines have the same gradient, but not the same intercept; do not keep the −5 from the first line.

+ Add to test✓ In your test54IGMC-043-00-07Gradient, midpoint and length of a line segment Paper CMedium9 marks

No calculator

A triangle has vertices A(1, 1), B(7, 3) and C(3, 7).

Diagram not to scale

(a)

Show that AB = AC.

(3)

(b)

M is the midpoint of BC. Find the coordinates of M.

(2)

(c)

Find the length AM. Leave your answer as a square root.

(2)

(d)

AM is perpendicular to BC. Find the area of triangle ABC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(AB^2 = 6^2+2^2\) or \(AC^2 = 2^2+6^2\)	✓ 1
A1 \(AB=\sqrt{40}\)	✓ 1
A1 \(AC=\sqrt{40}\), so AB = AC	✓ 1
Part (b)
M1 for \(\frac{7+3}{2}\) or \(\frac{3+7}{2}\)	✓ 1
A1 (5, 5)	✓ 1
Part (c)
M1 for \(\sqrt{4^2+4^2}\)	✓ 1
A1 \(\sqrt{32}\) or \(4\sqrt{2}\)	✓ 1
Part (d)
M1 for BC = \(\sqrt{4^2+4^2}=\sqrt{32}\) and \(\frac{1}{2}\times\sqrt{32}\times\sqrt{32}\)	✓ 1
A1 16	✓ 1

Exam tip: In an isosceles triangle, the line from the top vertex to the midpoint of the base is the perpendicular height, so you can use it directly in \(\frac{1}{2}\times\) base \(\times\) height.

+ Add to test✓ In your test55IGMC-043-00-09Gradient, midpoint and length of a line segment Paper CMedium4 marks

Calculator

A straight line passes through A(−2, −1) and B(5, 4).

Diagram not to scale

(a)

Find the gradient of AB.

(2)

(b)

Calculate the acute angle that AB makes with the positive x-axis. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{4-(-1)}{5-(-2)}\)	✓ 1
A1 \(\frac{5}{7}\)	✓ 1
Part (b)
M1 for \(\tan\theta=\frac{5}{7}\)	✓ 1
A1 35.5°	✓ 1

Exam tip: The gradient is the rise over the run, which is the tangent of the angle with the horizontal. Check your calculator is in degree mode.

+ Add to test✓ In your test56IGMC-043-00-10Gradient, midpoint and length of a line segment Paper CMedium5 marks

Calculator

On a map grid, each unit represents 1 km. A town is at the point (2, 3) and a village is at the point (14, 8). A cyclist travels in a straight line from the town to the village at a constant speed of 15 km/h.

(a)

Calculate the straight-line distance from the town to the village.

(3)

(b)

Calculate the time taken for the journey. Give your answer in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for differences 12 and 5	✓ 1
M1 for \(\sqrt{12^2+5^2}\)	✓ 1
A1 13 km	✓ 1
Part (b)
M1 for \(\frac{13}{15}\) or 0.866... hours	✓ 1
A1 52 minutes	✓ 1

Common error: 0.87 hours is not 87 minutes. Multiply the time in hours by 60 to convert it to minutes.

+ Add to test✓ In your test57IGMC-043-01-07Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

Calculator

The cost, C dollars, of a taxi journey of x kilometres follows a straight-line rule.
A 4 km journey costs 9.80 dollars and a 10 km journey costs 18.80 dollars.

(a)

Find the gradient of the line connecting C and x.

(2)

(b)

Find the equation connecting C and x, in the form C = mx + c.

(2)

(c)

Explain what the value of c represents in this context.

(1)

(d)

Calculate the cost of a 17 km journey.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(\frac{18.80-9.80}{10-4}\) or equivalent	✓ 1
A1: 1.5	✓ 1
Part (b)
M1: substitutes a point into C = 1.5x + c, e.g. 9.80 = 1.5 × 4 + c or 18.80 = 1.5 × 10 + c	✓ 1
A1: C = 1.5x + 3.8	✓ 1
Part (c)
B1: the fixed (starting) charge of 3.80 dollars, paid before any distance is travelled	✓ 1
Part (d)
B1: 29.30 dollars	✓ 1

Exam tip: In a real-life linear model the gradient is a rate (here dollars per kilometre) and the intercept is the value when x = 0.

+ Add to test✓ In your test58IGMC-043-01-11Equation of a straight line, parallel and perpendicular lines Paper CMedium4 marks

No calculator

The line \(y = -2x + c\) passes through the point (3, 1).

(a)

Find the value of c.

(2)

(b)

Find the coordinates of the point where the line crosses the x-axis.

(1)

(c)

Determine whether the point (−1, 9) lies on the line. Show your working.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 1 = −2 × 3 + c	✓ 1
A1: c = 7	✓ 1
Part (b)
B1: (3.5, 0)	✓ 1
Part (c)
B1: −2 × (−1) + 7 = 9, so yes, it lies on the line	✓ 1

Common error: On the x-axis it is y that equals 0, not x. Setting x = 0 gives the y-intercept instead.

+ Add to test✓ In your test59IGMC3-3-09c-07Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

A wheelchair ramp is modelled as a straight line on a scale drawing, in metres. It starts at ground level at A(1, 0) and ends on a platform at B(10, 0.75).

(a)

Find the gradient of the ramp as a fraction.

(2)

(b)

A guideline says a ramp must not be steeper than 1/15. Does this ramp meet the guideline? Show your working.

(2)

(c)

Calculate the length of the ramp surface.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.75 ÷ 9	✓ 1
A1 1/12	✓ 1
Part (b)
M1 compares 1/12 (0.0833) with 1/15 (0.0667)	✓ 1
A1 No, because 1/12 > 1/15 so it is too steep	✓ 1
Part (c)
M1 horizontal 9 and vertical 0.75	✓ 1
M1 √(9² + 0.75²) = √81.5625	✓ 1
A1 9.03 m	✓ 1

Common error: A larger denominator means a gentler slope: 1/15 is less steep than 1/12. Compare as decimals if unsure.

+ Add to test✓ In your test60IGMC3-3-09c-13Gradient, midpoint and length of a line segment Paper CMedium8 marks

Calculator

A robotic arm moves its gripper in a straight line from A(2, 1) to B(k, 7), where k > 2 and the units are centimetres. The distance AB is 10 cm.

(a)

Find the value of k.

(4)

(b)

Find the midpoint of AB.

(2)

(c)

Find the gradient of AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (k − 2)² + (7 − 1)² = 10²	✓ 1
M1 (k − 2)² = 64	✓ 1
A1 k − 2 = 8 (or −8)	✓ 1
A1 k = 10 since k > 2	✓ 1
Part (b)
B1 x = 6	✓ 1
B1 y = 4, so (6, 4)	✓ 1
Part (c)
M1 6/8	✓ 1
A1 3/4 (0.75)	✓ 1

Common error: Taking the square root gives two values; reject the one that breaks the condition in the question.

+ Add to test✓ In your test61IGMC3-3-09c-14Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

Calculator

On a plan of a bike track, a straight ramp lies along a line L that passes through (−2, 9) and has gradient −3/2.

(a)

Find the equation of L in the form y = mx + c.

(2)

(b)

Find the coordinates where L crosses the x-axis and the y-axis.

(2)

(c)

Calculate the distance between these two points.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes, e.g. 9 = −1.5(−2) + c	✓ 1
A1 y = −1.5x + 6	✓ 1
Part (b)
B1 (4, 0)	✓ 1
B1 (0, 6)	✓ 1
Part (c)
M1 √(4² + 6²) = √52	✓ 1
A1 7.21	✓ 1

Exam tip: For the x-intercept set y = 0; for the y-intercept set x = 0.

+ Add to test✓ In your test62IGMC3-3-09r5-05Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

Calculator

A service corridor in a hospital is modelled by a straight line through the points (2, 5) and (8, −1).

(a)

Calculate the gradient of the corridor line.

(2)

(b)

Find the equation of the line.

(2)

(c)

The point (k, 3) lies on the line. Find k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (−1 − 5)/(8 − 2)	✓ 1
A1 −1	✓ 1
Part (b)
M1 use y − 5 = −1(x − 2) or y = −x + c with a point substituted	✓ 1
A1 y = −x + 7	✓ 1
Part (c)
M1 3 = −k + 7	✓ 1
A1 k = 4	✓ 1

Common error: A negative gradient is easily lost when subtracting; keep the order of the coordinates the same in numerator and denominator.

+ Add to test✓ In your test63IGMC3-3-09r5-06Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

On a plan of a hospital car park, point M(4, 1) is the midpoint of a drainage channel AB. The end A is at (−2, 5).

Diagram not to scale

(a)

Find the coordinates of B.

(2)

(b)

Find the gradient of AB.

(2)

(c)

Calculate the length AB, giving your answer to 3 significant figures. One unit is 1 metre.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (−2 + x)/2 = 4 or x = 2 × 4 − (−2)	✓ 1
A1 B(10, −3)	✓ 1
Part (b)
M1 (−3 − 5)/(10 − (−2))	✓ 1
A1 −2/3 (accept −0.667)	✓ 1
Part (c)
M1 √(12² + 8²) (ECF from (a))	✓ 1
A1 √208 or 14.42…	✓ 1
A1 14.4 m	✓ 1

Exam tip: If the midpoint and one end are known, double the midpoint coordinates and subtract the known end.

+ Add to test✓ In your test64IGMC3-3-09r5-07Gradient, midpoint and length of a line segment Paper CMedium6 marks

Calculator

Three marshal posts on a Formula 1 circuit plan are at A(1, 1), B(7, 9) and C(13, 1). One unit on the plan represents 50 m.

Diagram not to scale

(a)

Show that the length AB is 10 units.

(2)

(b)

Show that triangle ABC is isosceles.

(2)

(c)

Find the perimeter of triangle ABC in metres.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(6² + 8²)	✓ 1
A1 √100 = 10	✓ 1
Part (b)
M1 BC = √(6² + 8²)	✓ 1
A1 BC = 10 = AB, so two sides are equal	✓ 1
Part (c)
B1 AC = 12 units	✓ 1
A1 (10 + 10 + 12) × 50 = 1600 m	✓ 1

Exam tip: When a scale is given, finish the geometry in plan units first and apply the scale once at the end.

+ Add to test✓ In your test65IGMC3-3-09r5-08Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

Calculator

A path past some beehives is modelled by a straight line through the points (4, k) and (8, 13). The gradient of the path is 2.5.

(a)

Find the value of k.

(3)

(b)

Find the equation of the path in the form y = mx + c.

(2)

(c)

Write down the y-intercept of the path.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (13 − k)/(8 − 4) = 2.5	✓ 1
M1 13 − k = 10	✓ 1
A1 k = 3	✓ 1
Part (b)
M1 y − 13 = 2.5(x − 8) or c found by substitution	✓ 1
A1 y = 2.5x − 7	✓ 1
Part (c)
B1 -7 (ECF from (b))	✓ 1

Common error: Here the unknown is a coordinate, not the gradient: set the gradient formula equal to the given value and solve.

+ Add to test✓ In your test66IGMC3-3-09r8-05Gradient, midpoint and length of a line segment Paper CMedium5 marks

Calculator

In a warehouse plan with units in metres, a rail runs in a straight line from A(2, k) to B(7, 9). The gradient of the rail is 2.

(a)

Find the value of k.

(3)

(b)

Calculate the length of the rail AB. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (9 − k)/(7 − 2) = 2	✓ 1
M1 9 − k = 10	✓ 1
A1 k = −1	✓ 1
Part (b)
M1 √((7 − 2)² + (9 − (−1))²) = √125	✓ 1
A1 11.2 m	✓ 1

Exam tip: Brackets matter when a coordinate is negative: (9 − (−1)) = 10, not 8.

+ Add to test✓ In your test67IGMC3-3-09r8-06Gradient, midpoint and length of a line segment Paper CMedium5 marks

No calculator

In an architect's elevation drawing the edge of a ramp lies on the line y = 3x − 6, with units in metres.

(a)

Write down the y-intercept of the line.

(1)

(b)

Find the x-intercept of the line.

(2)

(c)

The line and the two axes enclose a triangle. Find its area.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 (0, −6) or −6	✓ 1
Part (b)
M1 3x − 6 = 0	✓ 1
A1 x = 2	✓ 1
Part (c)
M1 ½ × 2 × 6	✓ 1
A1 6 m²	✓ 1

Common error: Using −6 as a length for the base or height; lengths in an area are positive.

+ Add to test✓ In your test68IGMC3-3-09r8-07Equation of a straight line, parallel and perpendicular lines Paper CMedium5 marks

No calculator

On a reef survey grid, a transect line passes through the point (3, −2) and has gradient ½.

(a)

Find the equation of the transect line in the form y = mx + c.

(3)

(b)

A marker lies on the line at y = 4. Find its x-coordinate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes (3, −2) and ½ into y = mx + c	✓ 1
A1 c = −3.5	✓ 1
A1 y = 0.5x − 3.5	✓ 1
Part (b)
M1 4 = 0.5x − 3.5	✓ 1
A1 x = 15	✓ 1

Exam tip: Substitute a given point into your final equation as a check before moving on.

+ Add to test✓ In your test69IGMC3-3-09r8-08Gradient, midpoint and length of a line segment Paper CMedium5 marks

No calculator

Three beacons stand at A(1, 2), B(4, 8) and C(6, 12) on a harbour chart.

(a)

Show that A, B and C lie on a straight line.

(3)

(b)

The line through the beacons has equation y = 2x. A fourth beacon D(10, k) lies on the same line. Find k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient AB = (8 − 2)/(4 − 1) = 2	✓ 1
M1 gradient BC = (12 − 8)/(6 − 4) = 2	✓ 1
R1 equal gradients and the common point B, so A, B, C are collinear	✓ 1
Part (b)
M1 k = 2 × 10	✓ 1
A1 k = 20	✓ 1

Common error: Comparing only one gradient; both gradients, sharing a point, are needed to prove collinearity.

+ Add to test✓ In your test70IGMC3-3-09r8-14Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

A tagged seal is recorded at A(1, 2) and later at B(7, k) on a grid with units of 1 km, where k > 0. The distance AB is 10 km.

(a)

Show that (k − 2)² = 64.

(2)

(b)

Find the value of k.

(2)

(c)

M is the midpoint of AB. Calculate the distance from M to the origin O(0, 0). Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (7 − 1)² + (k − 2)² = 10²	✓ 1
A1 36 + (k − 2)² = 100, so (k − 2)² = 64	✓ 1
Part (b)
M1 k − 2 = ±8	✓ 1
A1 k = 10 (k = −6 rejected as k > 0)	✓ 1
Part (c)
B1 M = (4, 6)	✓ 1
M1 √(4² + 6²) = √52	✓ 1
A1 7.21 km	✓ 1

Common error: Dropping the negative root when solving a squared equation; use the condition given to choose between the roots.

+ Add to test✓ In your test71IGMC3-3-09r12-05Gradient, midpoint and length of a line segment Paper CMedium8 marks

Calculator

A research vessel sails along a straight line that passes through the points (−4, 7) and (5, −2) on a map grid. Each unit represents 1 km.

(a)

Find the gradient of the line.

(2)

(b)

Find the equation of the line in the form y = mx + c.

(2)

(c)

A marker buoy is at B(10, −8). Show that B does not lie on the line.

(2)

(d)

Calculate the distance between the two given points, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{-2-7}{5-(-4)}\)	✓ 1
A1 −1	✓ 1
Part (b)
M1 substitute a point and the gradient into y = mx + c	✓ 1
A1 y = −x + 3	✓ 1
Part (c)
M1 substitute x = 10 to get y = −7	✓ 1
A1 −7 ≠ −8, so B is not on the line	✓ 1
Part (d)
M1 \(\sqrt{9^2+9^2}\) or \(\sqrt{162}\)	✓ 1
A1 12.7 km	✓ 1

Exam tip: For a 'show that' part, state the value you obtain and compare it with the given value in a closing sentence.

+ Add to test✓ In your test72IGMC3-3-09r12-07Gradient, midpoint and length of a line segment Paper CMedium5 marks

Calculator

A delivery route runs in a straight line from A(−6, −3) to B(10, 9). Each unit on the grid represents 1 km. A refuelling point is at the midpoint of AB.

Diagram not to scale

(a)

Find the coordinates of the refuelling point.

(2)

(b)

A lorry drives from A to B at a constant speed of 40 km/h. Calculate the time taken, in minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\left(\frac{-6+10}{2},\frac{-3+9}{2}\right)\)	✓ 1
A1 (2, 3)	✓ 1
Part (b)
M1 length \(\sqrt{16^2+12^2}\)	✓ 1
A1 20 km	✓ 1
A1 30 minutes	✓ 1

Exam tip: Convert hours to minutes only at the end: 0.5 hours is 30 minutes.

+ Add to test✓ In your test73IGMC3-3-09r12-10Equation of a straight line, parallel and perpendicular lines Paper CMedium5 marks

Calculator

A cable is to be laid in a straight line from A(1, 2) to B(p, 10), where p > 1. Each unit on the grid represents 1 m, and the length AB is 10 m.

(a)

Find the value of p.

(3)

(b)

The cable costs €12.50 per metre. Calculate the total cost of the cable.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \((p-1)^2+(10-2)^2=10^2\)	✓ 1
A1 \((p-1)^2=36\)	✓ 1
A1 p = 7 (the value −5 is rejected as p > 1)	✓ 1
Part (b)
M1 10 × 12.50	✓ 1
A1 €125.00	✓ 1

Exam tip: When squaring gives two solutions, use the condition in the question to reject the one that does not fit.

+ Add to test✓ In your test74IGMC3-3-09r12-11Equation of a straight line, parallel and perpendicular lines Paper CMedium7 marks

No calculator

A line L in a coral survey passes through A(0, −6) and B(2, 2). Each unit represents 1 m.

(a)

Find the equation of L in the form y = mx + c.

(3)

(b)

Does the point C(3, 5) lie on L? Show how you decide.

(2)

(c)

Find the x-coordinate of the point on L where y = 10.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient \(\frac{2-(-6)}{2-0}\)	✓ 1
A1 m = 4	✓ 1
A1 y = 4x − 6 (A is the y-intercept)	✓ 1
Part (b)
M1 substitute x = 3 to get y = 6	✓ 1
A1 6 ≠ 5, so C is not on L	✓ 1
Part (c)
M1 10 = 4x − 6	✓ 1
A1 x = 4	✓ 1

Exam tip: A point lies on a line only if its coordinates make the equation true.

+ Add to test✓ In your test75IGMC3-3-10r15-05Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

On a map of a water-supply network, a reservoir is at R(1, 2) and a town is at T(8, −5). Coordinates are in kilometres. A straight pipe is to be laid from R to T.

(a)

A pumping station is to be built at the midpoint of RT. Find its coordinates.

(2)

(b)

Calculate the length of the pipe, to 3 significant figures.

(3)

(c)

Laying the pipe costs €240 for each kilometre. Using the unrounded length from (b), calculate the cost to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ((1 + 8)/2, (2 + (−5))/2)	✓ 1
A1 (4.5, −1.5)	✓ 1
Part (b)
M1 (8 − 1)² + (−5 − 2)² = 98	✓ 1
M1 √98	✓ 1
A1 9.90 km	✓ 1
Part (c)
M1 9.8994… × 240	✓ 1
A1 €2376	✓ 1

Common error: Both coordinate differences are squared before adding; squaring a negative difference gives a positive value.

+ Add to test✓ In your test76IGMC3-3-10r15-06Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

No calculator

In a game, a path starts at A(1, 4) and ends at B(7, k). The gradient of the path AB is ½.

(a)

Find the value of k.

(2)

(b)

Find the equation of the line AB in the form y = mx + c.

(2)

(c)

A character at (−5, p) lies on the line AB. Find p.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (k − 4)/(7 − 1) = ½	✓ 1
A1 k = 7	✓ 1
Part (b)
M1 y − 4 = ½(x − 1), or 4 = ½ × 1 + c	✓ 1
A1 y = ½x + 3.5	✓ 1
Part (c)
M1 p = ½ × (−5) + 3.5	✓ 1
A1 p = 1	✓ 1

Exam tip: After finding an equation, check it by substituting a known point such as B(7, 7): ½ × 7 + 3.5 = 7.

+ Add to test✓ In your test77IGMC3-3-10r15-07Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

A water pipe has one end at P(−2, 4). The midpoint of the pipe is M(3, −1). The other end is at Q.

(a)

Find the coordinates of Q.

(3)

(b)

Find the gradient of PQ.

(2)

(c)

Calculate the length of the pipe PQ, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (−2 + x)/2 = 3 or (4 + y)/2 = −1	✓ 1
A1 x = 8	✓ 1
A1 y = −6, so Q(8, −6)	✓ 1
Part (b)
M1 (−6 − 4)/(8 − (−2))	✓ 1
A1 −1	✓ 1
Part (c)
M1 (8 − (−2))² + (−6 − 4)² = 200	✓ 1
A1 14.1 units	✓ 1

Exam tip: To find an end point from a midpoint, double the midpoint's coordinates and subtract the known end's coordinates.

+ Add to test✓ In your test78IGMC3-3-10r15-08Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

The dose D mg of a medicine depends on a child's mass w kg. The relationship is linear. A child of mass 20 kg is given 150 mg and a child of mass 60 kg is given 350 mg.

(a)

Calculate the gradient of the line.

(2)

(b)

Find the equation of the line, giving D in terms of w.

(2)

(c)

State what the gradient represents in this context.

(1)

(d)

A child is given a dose of 400 mg. Find the mass of the child.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (350 − 150)/(60 − 20)	✓ 1
A1 5	✓ 1
Part (b)
M1 150 = 5 × 20 + c	✓ 1
A1 D = 5w + 50	✓ 1
Part (c)
B1 each extra kilogram of mass adds 5 mg to the dose	✓ 1
Part (d)
M1 400 = 5w + 50	✓ 1
A1 w = 70 kg	✓ 1

Common error: Substitute the value into the correct variable: here D is 400, not w.

+ Add to test✓ In your test79IGMC3-3-10r15-14Gradient, midpoint and length of a line segment Paper CMedium7 marks

No calculator

On a game map, a player is at A(1, 2) and an enemy is at B(k, 8), where k > 1. The distance AB is 10 units.

(a)

Find the value of k.

(3)

(b)

Find the midpoint of AB.

(2)

(c)

Find the gradient of AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (k − 1)² + (8 − 2)² = 10²	✓ 1
A1 (k − 1)² = 64	✓ 1
A1 k = 9 (since k > 1)	✓ 1
Part (b)
M1 ((1 + 9)/2, (2 + 8)/2)	✓ 1
A1 (5, 5)	✓ 1
Part (c)
M1 (8 − 2)/(9 − 1)	✓ 1
A1 3/4 (= 0.75)	✓ 1

Exam tip: When solving (k − 1)² = 64, take the square root and use the condition on k to reject the negative value.

+ Add to test✓ In your test80IGMC3-3-10r21-07Gradient, midpoint and length of a line segment Paper CMedium6 marks

Calculator

Three mountaineering camps are at A(−1, 2), B(5, 4) and C(3, −2) on a map grid, with units in kilometres.

Diagram not to scale

(a)

Show that AB = BC by finding both lengths.

(3)

(b)

State the special name of triangle ABC.

(1)

(c)

Calculate the perimeter of the triangle, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(AB=\sqrt{(5+1)^2+(4-2)^2}\)	✓ 1
M1 \(BC=\sqrt{(3-5)^2+(-2-4)^2}\)	✓ 1
A1 both equal \(\sqrt{40}\)	✓ 1
Part (b)
B1 isosceles	✓ 1
Part (c)
M1 \(AC=\sqrt{(3+1)^2+(-2-2)^2}=\sqrt{32}\) and \(2\sqrt{40}+\sqrt{32}\)	✓ 1
A1 18.3 km	✓ 1

Common error: Forgetting to square negative differences correctly: \((-2)^2=4\), not −4.

+ Add to test✓ In your test81IGMC3-3-10r21-10Gradient, midpoint and length of a line segment Paper CMedium5 marks

No calculator

On a map grid in kilometres, mast A is at (2, 5) and mast B is at (k, 11), where k > 0. The distance AB is 10 km.

(a)

Show that \((k-2)^2=64\).

(2)

(b)

Find the value of k.

(2)

(c)

Find the gradient of AB.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \((k-2)^2+(11-5)^2=10^2\)	✓ 1
A1 \((k-2)^2+36=100\), so \((k-2)^2=64\)	✓ 1
Part (b)
M1 \(k-2=\pm8\)	✓ 1
A1 \(k=10\) (since \(k>0\); \(k=-6\) rejected)	✓ 1
Part (c)
A1 \(\frac{6}{8}=0.75\)	✓ 1

Common error: Giving only one value from a square root. Write both \(\pm\) roots and then use the condition in the question to reject one.

+ Add to test✓ In your test82IGMC3-3-10r21-11Gradient, midpoint and length of a line segment Paper CMedium6 marks

Calculator

A cable car line is drawn on a graph with horizontal distance d km and height h km. Its lower station is at (2, 0.8) and its upper station is at (8, 2.3).

(a)

Calculate the gradient of the cable car line.

(2)

(b)

Calculate the length of the cable between the stations, to 3 significant figures.

(2)

(c)

Find the equation of the line in the form \(h=md+c\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{2.3-0.8}{8-2}\)	✓ 1
A1 0.25	✓ 1
Part (b)
M1 \(\sqrt{(8-2)^2+(2.3-0.8)^2}\)	✓ 1
A1 6.18 km	✓ 1
Part (c)
M1 \(0.8=0.25\times2+c\)	✓ 1
A1 \(h=0.25d+0.3\)	✓ 1

Exam tip: The cable length is the hypotenuse of the right-angled triangle with the horizontal and vertical differences as its shorter sides.

+ Add to test✓ In your test83IGMC3-3-11r29-05Gradient, midpoint and length of a line segment Paper CMedium8 marks

Calculator

Three inverters at a solar farm are positioned at A(1, 2), B(7, 10) and C(13, 2). The coordinates are in metres.

Diagram not to scale

(a)

Calculate the length of AB.

(2)

(b)

Show that triangle ABC is isosceles.

(2)

(c)

Find the coordinates of the midpoint M of AC.

(1)

(d)

Calculate the area of triangle ABC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(6² + 8²)	✓ 1
A1 10 m	✓ 1
Part (b)
M1 BC = √(6² + 8²)	✓ 1
A1 BC = 10 = AB, so two sides are equal	✓ 1
Part (c)
B1 (7, 2)	✓ 1
Part (d)
B1 AC = 12 and BM = 8 (BM is perpendicular to AC)	✓ 1
M1 ½ × 12 × 8	✓ 1
A1 48 m²	✓ 1

Exam tip: In an isosceles triangle the line from the apex to the midpoint of the base is perpendicular to it, which gives the height directly.

+ Add to test✓ In your test84IGMC3-3-11r29-08Equation of a straight line, parallel and perpendicular lines Paper CMedium8 marks

Calculator

In a game, a character walks in a straight line from P(−2, 7) to Q(4, −5).

(a)

Calculate the gradient of PQ.

(2)

(b)

Find the equation of the line PQ in the form y = mx + c.

(2)

(c)

The point R(k, −9) lies on PQ. Find k.

(2)

(d)

Find the x-coordinate of the point where the path crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (−5 − 7)/(4 − (−2))	✓ 1
A1 −2	✓ 1
Part (b)
M1 y − 7 = −2(x + 2)	✓ 1
A1 y = −2x + 3	✓ 1
Part (c)
M1 −9 = −2k + 3	✓ 1
A1 k = 6	✓ 1
Part (d)
M1 0 = −2x + 3	✓ 1
A1 1.5	✓ 1

Exam tip: Test your equation with the second point: 3 − 2(4) = −5, which matches Q.

+ Add to test✓ In your test85IGMC3-3-11r29-10Gradient, midpoint and length of a line segment Paper CMedium6 marks

Calculator

A medicine-delivery drone flies from a depot at A(−4, 3) to a hospital at B(8, −2), where 1 unit represents 1 km. It then continues in the same straight line to a second clinic at C, so that B is the midpoint of AC.

Diagram not to scale

(a)

Show that the distance AB is 13 km.

(2)

(b)

Find the coordinates of C.

(2)

(c)

Calculate the distance AC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(12² + 5²)	✓ 1
A1 √169 = 13	✓ 1
Part (b)
M1 (−4 + x)/2 = 8 and (3 + y)/2 = −2	✓ 1
A1 C = (20, −7)	✓ 1
Part (c)
M1 √(24² + 10²) (or 2 × 13)	✓ 1
A1 26 km	✓ 1

Exam tip: If B is the midpoint of AC, then AC = 2 × AB, which gives a quick check on your answer.

+ Add to test✓ In your test86IGMC3-3-11r29-11Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

A robot moves around a hospital ward plan where 1 unit represents 1 metre. The robot starts at A(−3, k) and stops at B(6, 9). The line AB has gradient 2/3.

(a)

Show that k = 3.

(3)

(b)

Find the coordinates of the midpoint of AB.

(2)

(c)

Calculate the distance the robot travels, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (9 − k)/(6 − (−3)) = 2/3	✓ 1
M1 9 − k = 6	✓ 1
A1 k = 3	✓ 1
Part (b)
M1 attempt to average coordinates	✓ 1
A1 (1.5, 6)	✓ 1
Part (c)
M1 √(9² + 6²)	✓ 1
A1 10.8 m	✓ 1

Exam tip: When a gradient is given, write it as an equation using the coordinates and solve for the unknown instead of guessing.

+ Add to test✓ In your test87IGMC3-3-11r35-05Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

On a navigation map, 1 unit represents 1 km. Two beacons are at A(−4, 1) and B(8, −4).

Diagram not to scale

(a)

Calculate the distance between the beacons.

(3)

(b)

Find the coordinates of the midpoint of AB.

(2)

(c)

A light aircraft flies directly from A to B at a constant speed of 260 km/h. Calculate the time taken, in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differences 12 and 5 used correctly	✓ 1
M1 √(12² + 5²) = √169	✓ 1
A1 13 km	✓ 1
Part (b)
B1 x = 2	✓ 1
B1 y = −1.5	✓ 1
Part (c)
M1 13 ÷ 260 = 0.05 hours (ECF from (a))	✓ 1
A1 3 minutes	✓ 1

Common error: Convert time units at the end: 0.05 hours is 0.05 × 60 minutes, not 5 minutes.

+ Add to test✓ In your test88IGMC3-3-11r35-06Gradient, midpoint and length of a line segment Paper CMedium8 marks

Calculator

On a plan of a farm, each unit represents 10 m. A triangular field has corners A(0, 0), B(6, 8) and C(12, 0).

Diagram not to scale

(a)

Calculate the length of AB, in units.

(2)

(b)

Show that BC has the same length as AB.

(2)

(c)

Fencing costs £4.50 per metre. Calculate the cost of fencing the whole perimeter of the field.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(6² + 8²)	✓ 1
A1 10	✓ 1
Part (b)
M1 √((12 − 6)² + (0 − 8)²) = √(36 + 64)	✓ 1
A1 = 10, so BC = AB	✓ 1
Part (c)
B1 AC = 12 units	✓ 1
M1 perimeter (10 + 10 + 12) × 10 = 320 m	✓ 1
M1 320 × 4.50	✓ 1
A1 £1440	✓ 1

Exam tip: Read the scale carefully: lengths found from coordinates are in units and must be converted to metres before costing.

+ Add to test✓ In your test89IGMC3-3-11r35-07Equation of a straight line, parallel and perpendicular lines Paper CMedium7 marks

Calculator

A sound engineer draws a calibration line for an amplifier. The line passes through A(1, k) and B(5, 11) and has gradient 2.5.

(a)

Find the value of k.

(3)

(b)

Find the equation of the line in the form y = mx + c.

(2)

(c)

The line crosses the x-axis at C. Find the x-coordinate of C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (11 − k) ÷ (5 − 1) = 2.5	✓ 1
M1 11 − k = 10	✓ 1
A1 k = 1	✓ 1
Part (b)
M1 y = 2.5x + c with a correct substitution, e.g. 11 = 12.5 + c	✓ 1
A1 y = 2.5x − 1.5	✓ 1
Part (c)
M1 0 = 2.5x − 1.5 (ECF from (b))	✓ 1
A1 0.6	✓ 1

Common error: In the gradient formula keep the coordinates in matching order; (11 − k) ÷ (5 − 1) and (k − 11) ÷ (1 − 5) are equivalent, but mixing them gives the wrong sign.

+ Add to test✓ In your test90IGMC3-3-11r35-10Gradient, midpoint and length of a line segment Paper CMedium6 marks

Calculator

A stage plan uses a grid in which 1 unit represents 1 m. A speaker is at P(−6, −2) and another speaker is at Q(6, 7). A cable runs in a straight line from P to Q.

Diagram not to scale

(a)

Find the gradient of PQ.

(2)

(b)

A microphone stand is at R(2, 4). Show that R lies on the cable PQ.

(2)

(c)

Calculate the length of the cable.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (7 − (−2)) ÷ (6 − (−6))	✓ 1
A1 3/4 (or 0.75)	✓ 1
Part (b)
M1 gradient of PR = (4 − (−2)) ÷ (2 − (−6)) = 6/8	✓ 1
A1 = 0.75, the same as PQ, and P is common to both, so R lies on PQ	✓ 1
Part (c)
M1 √(12² + 9²)	✓ 1
A1 15 m	✓ 1

Exam tip: To show that three points lie on a straight line, find two gradients from a common point and state that they are equal.

+ Add to test✓ In your test91IGMC3-3-11r35-11Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

No calculator

Two stakes in a pasture are at A(−2, 7) and B(4, −5), on a plan with 1 unit representing 1 m. A rope is stretched along the line through A and B.

(a)

Find the gradient of the rope.

(2)

(b)

Find the equation of the line AB in the form y = mx + c.

(2)

(c)

The line AB crosses the x-axis at D. Find the x-coordinate of D.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (−5 − 7) ÷ (4 − (−2))	✓ 1
A1 −2	✓ 1
Part (b)
M1 y = −2x + c with a correct substitution, e.g. 7 = 4 + c	✓ 1
A1 y = −2x + 3	✓ 1
Part (c)
M1 0 = −2x + 3 (ECF from (b))	✓ 1
A1 1.5	✓ 1

Common error: A negative gradient means the line goes down as you move right; check that your sign agrees with this.

+ Add to test✓ In your test92IGMC3-3-10r43-03Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

Calculator

The cost, y euros, of making x dresses in a small fashion workshop is modelled by a straight line. The points (10, 270) and (30, 510) lie on this line.

(a)

Find the gradient of the line.

(2)

(b)

Find the equation of the line in the form y = mx + c.

(2)

(c)

Write down what the gradient represents in this situation.

(1)

(d)

Use the model to find the cost of making 45 dresses.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (510 − 270) / (30 − 10)	✓ 1
A1 12	✓ 1
Part (b)
M1 y − 270 = 12(x − 10) or 270 = 12 × 10 + c	✓ 1
A1 y = 12x + 150	✓ 1
Part (c)
B1 the extra cost (€12) of making each additional dress	✓ 1
Part (d)
B1 €690 (12 × 45 + 150; ECF from (b))	✓ 1

Exam tip: The gradient of a cost line is the cost per extra item, and the intercept is the fixed cost.

+ Add to test✓ In your test93IGMC3-3-10r43-04Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

No calculator

The path of a robot is a straight line through the points P(1, 5) and Q(4, 11).

(a)

Find the gradient of PQ.

(2)

(b)

Find the equation of the line PQ in the form y = mx + c.

(2)

(c)

Find the x-coordinate of the point where the path crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (11 − 5) / (4 − 1)	✓ 1
A1 2	✓ 1
Part (b)
M1 y − 5 = 2(x − 1) or 5 = 2 × 1 + c	✓ 1
A1 y = 2x + 3	✓ 1
Part (c)
M1 0 = 2x + 3	✓ 1
A1 x = −3/2 (= −1.5)	✓ 1

Common error: On the x-axis the y-coordinate is zero; substitute y = 0, not x = 0.

+ Add to test✓ In your test94IGMC3-3-10r43-05Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

Calculator

A fishing net is stretched along the line with equation 3x + 4y = 24. Coordinates are in metres and the net runs between the two axes.

(a)

Rearrange the equation to find the gradient of the line.

(2)

(b)

Find the coordinates of the points where the line meets the x-axis and the y-axis.

(2)

(c)

Calculate the length of the net between the two axes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4y = −3x + 24 (or y = −(3/4)x + 6)	✓ 1
A1 gradient = −3/4 (−0.75)	✓ 1
Part (b)
B1 (8, 0)	✓ 1
B1 (0, 6)	✓ 1
Part (c)
M1 √(8² + 6²)	✓ 1
A1 10 m	✓ 1

Exam tip: To find intercepts, set one variable to zero in turn and solve for the other.

+ Add to test✓ In your test95IGMC3-3-10r43-11Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

A triangular patch of fabric ABC has vertices A(1, 1), B(5, 4) and C(−2, 5), with all lengths in centimetres.

Diagram not to scale

(a)

Calculate the length of AB.

(2)

(b)

Show that triangle ABC is isosceles.

(2)

(c)

Calculate the perimeter of the triangle. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(4² + 3²)	✓ 1
A1 5 cm	✓ 1
Part (b)
M1 AC = √((−2 − 1)² + (5 − 1)²)	✓ 1
A1 AC = 5 = AB, so two sides are equal	✓ 1
Part (c)
M1 BC = √((−2 − 5)² + (5 − 4)²)	✓ 1
A1 BC = √50 = 7.07 (7.0711…)	✓ 1
A1 perimeter = 17.1 cm	✓ 1

Exam tip: In a perimeter question, keep the unrounded value of each side until the final addition.

+ Add to test✓ In your test96IGMC3-3-11r44-05Gradient, midpoint and length of a line segment Paper CMedium8 marks

Calculator

A mobile testing van drives along a straight road on a map grid in which 1 unit represents 1 km. The road passes through the points (−4, 7) and (2, −2).

(a)

Find the equation of the road in the form y = mx + c.

(3)

(b)

Show that the point (6, −8) lies on the road.

(2)

(c)

Calculate the distance along the road between the points (−4, 7) and (2, −2). Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient (−2 − 7)/(2 − (−4))	✓ 1
A1 −1.5 (accept −3/2)	✓ 1
A1 y = −1.5x + 1	✓ 1
Part (b)
M1 substitutes x = 6 to get −1.5(6) + 1	✓ 1
A1 = −8, so the point is on the road	✓ 1
Part (c)
M1 correct differences 6 and 9	✓ 1
M1 √(6² + 9²) or √117	✓ 1
A1 10.8 km	✓ 1

Common error: Subtract the coordinates in the same order in the numerator and the denominator when finding a gradient. Reversing one of them changes the sign.

+ Add to test✓ In your test97IGMC3-3-11r44-06Gradient, midpoint and length of a line segment Paper CMedium6 marks

Calculator

On a map for a solar-farm project, 1 unit represents 1 km. A power line follows the straight line 3y + 2x = 12.

(a)

Find the gradient of the power line.

(2)

(b)

Find the coordinates of the points where the power line crosses the x-axis and the y-axis.

(2)

(c)

Calculate the distance between these two points. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges to y = −(2/3)x + 4	✓ 1
A1 −2/3	✓ 1
Part (b)
B1 (6, 0)	✓ 1
B1 (0, 4)	✓ 1
Part (c)
M1 √(6² + 4²) or √52	✓ 1
A1 7.21 km	✓ 1

Exam tip: To find axis crossings, set one variable to zero in turn. Check each point by substituting back into the original equation.

+ Add to test✓ In your test98IGMC3-3-11r44-07Gradient, midpoint and length of a line segment Paper CMedium6 marks

No calculator

The plan of a retail park uses coordinates in metres. Three stores are at A(−1, 2), B(7, 6) and C(3, −4). M is the midpoint of AB.

Diagram not to scale

(a)

Find the coordinates of M.

(2)

(b)

Find the gradient of AC.

(2)

(c)

Find the length of MC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (−1 + 7)/2 or (2 + 6)/2	✓ 1
A1 (3, 4)	✓ 1
Part (b)
M1 (−4 − 2)/(3 − (−1))	✓ 1
A1 −3/2	✓ 1
Part (c)
M1 correct differences 0 and 8, or recognises that MC is vertical	✓ 1
A1 8 m	✓ 1

Common error: Do not skip the check for special cases: when two points share an x-coordinate the line is vertical and the length is just the difference in y.

+ Add to test✓ In your test99IGMC3-3-11r44-08Equation of a straight line, parallel and perpendicular lines Paper CMedium7 marks

Calculator

Two screening centres are at P(−3, −1) and Q(9, 4) on a map grid where 1 unit represents 1 km.

Diagram not to scale

(a)

Show that the distance PQ is 13 km.

(2)

(b)

Find the equation of the line PQ. Give your answer in the form ay = bx + c, where a, b and c are integers.

(3)

(c)

The line PQ meets the x-axis at the point R. Find the x-coordinate of R.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(12² + 5²)	✓ 1
A1 √169 = 13 km	✓ 1
Part (b)
M1 gradient 5/12	✓ 1
M1 substitutes a point, e.g. y + 1 = (5/12)(x + 3)	✓ 1
A1 12y = 5x + 3 (or any equivalent correct form)	✓ 1
Part (c)
M1 sets y = 0	✓ 1
A1 x = −0.6 (accept −3/5)	✓ 1

Exam tip: When a question asks for integer coefficients, clear the fractions by multiplying through by the common denominator before writing the final equation.

+ Add to test✓ In your test100IGMC3-3-10r47-06Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

A cable car runs in a straight line from station S(0.5, 1.2) to station T(6.5, 3.6). Units are kilometres.

Diagram not to scale

(a)

Find the coordinates of the midpoint of ST, where a pylon is to be built.

(2)

(b)

Find the gradient of ST.

(2)

(c)

Calculate the length of ST. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ((0.5 + 6.5)/2, (1.2 + 3.6)/2)	✓ 1
A1 (3.5, 2.4)	✓ 1
Part (b)
M1 (3.6 − 1.2) / (6.5 − 0.5)	✓ 1
A1 0.4	✓ 1
Part (c)
M1 differences of 6 and 2.4	✓ 1
M1 length = √(6² + 2.4²) = √41.76	✓ 1
A1 6.46 km	✓ 1

Exam tip: Use the unrounded value of the length on your calculator and round only at the end.

+ Add to test✓ In your test101IGMC3-3-10r47-07Equation of a straight line, parallel and perpendicular lines Paper CMedium8 marks

Calculator

A cheese seller's price £C for a block of cheese of mass m kg is modelled by a straight line. A block of 0.5 kg costs £5.40 and a block of 2.5 kg costs £16.20.

(a)

Find the gradient of the line through (0.5, 5.40) and (2.5, 16.20).

(2)

(b)

Find the equation of the line in the form C = km + c.

(2)

(c)

Explain what the gradient and the value of c mean in this situation.

(2)

(d)

Use the model to find the cost of a block of mass 4.2 kg.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (16.20 − 5.40) / (2.5 − 0.5)	✓ 1
A1 5.4	✓ 1
Part (b)
M1 5.40 = 5.4 × 0.5 + c	✓ 1
A1 C = 5.4m + 2.7	✓ 1
Part (c)
B1 the gradient is the cost per kilogram, £5.40 for each kg of cheese	✓ 1
B1 c is a fixed charge of £2.70 that does not depend on the mass, for example packaging	✓ 1
Part (d)
M1 5.4 × 4.2 + 2.7 = 22.68 + 2.7	✓ 1
A1 £25.38	✓ 1

Exam tip: In a modelling question, give the meaning of the gradient with its units: pounds per kilogram.

+ Add to test✓ In your test102IGMC3-3-10r47-09Equation of a straight line, parallel and perpendicular lines Paper CMedium5 marks

No calculator

Three huts on a ridge are at P(0, 1), Q(4, 4) and R(8, 1). Units are kilometres.

Diagram not to scale

(a)

Show that the distance PQ equals the distance QR.

(3)

(b)

Find the total length of the route P to Q to R and back to P.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 PQ = √(4² + 3²)	✓ 1
M1 QR = √(4² + (−3)²)	✓ 1
A1 both equal 5, so PQ = QR	✓ 1
Part (b)
M1 PR = 8 and perimeter = 5 + 5 + 8 (ECF from (a))	✓ 1
A1 18 km	✓ 1

Exam tip: In a 'show that' question write each distance in full and finish with a clear statement that they are equal.

+ Add to test✓ In your test103IGMC3-3-10r47-11Equation of a straight line, parallel and perpendicular lines Paper CMedium6 marks

Calculator

A boundary fence follows the straight line 3x + 5y = 30. Units are kilometres.

(a)

Rearrange the equation to the form y = mx + c and state the gradient.

(2)

(b)

Find the coordinates of the points where the fence crosses the x-axis and the y-axis.

(2)

(c)

Calculate the length of fence between these two points. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5y = 30 − 3x, so y = −0.6x + 6	✓ 1
A1 gradient −0.6 (or −3/5)	✓ 1
Part (b)
B1 x-axis crossing (10, 0)	✓ 1
B1 y-axis crossing (0, 6)	✓ 1
Part (c)
M1 √(10² + 6²) = √136 (ECF from (b))	✓ 1
A1 11.7 km	✓ 1

Exam tip: To find where a line crosses an axis, set the other coordinate to zero.

+ Add to test✓ In your test104IGMC3-3-10r47-14Gradient, midpoint and length of a line segment Paper CMedium7 marks

Calculator

A telecoms cable runs in a straight line from A(2, 5) to B(k, 11). The gradient of AB is 1.5. Units are kilometres.

(a)

Find the value of k.

(3)

(b)

Find the coordinates of the midpoint of AB.

(2)

(c)

Calculate the length of AB. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (11 − 5)/(k − 2) = 1.5	✓ 1
M1 k − 2 = 6 / 1.5 = 4	✓ 1
A1 k = 6	✓ 1
Part (b)
M1 ((2 + 6)/2, (5 + 11)/2) (ECF from (a))	✓ 1
A1 (4, 8)	✓ 1
Part (c)
M1 √(4² + 6²) = √52 (ECF from (a))	✓ 1
A1 7.21 km	✓ 1

Common error: The gradient is change in y divided by change in x, so a larger k gives a smaller gradient; do not invert the fraction.

+ Add to test✓ In your test105IGMC3-3-10r21-14Gradient, midpoint and length of a line segment Paper CHard8 marks

No calculator

In a food science laboratory, four sample positions A(1, 1), B(6, 2), C(9, 6) and D are the vertices of a parallelogram ABCD, with units in centimetres. The diagonals of a parallelogram bisect each other.

Diagram not to scale

(a)

Find the midpoint of AC.

(2)

(b)

Use your answer to (a) to find the coordinates of D.

(2)

(c)

Show that AB is parallel to DC.

(3)

(d)

Find the length of AD.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\left(\frac{1+9}{2},\frac{1+6}{2}\right)\)	✓ 1
A1 (5, 3.5)	✓ 1
Part (b)
M1 midpoint of BD is (5, 3.5), so \(\frac{6+x}{2}=5\) and \(\frac{2+y}{2}=3.5\)	✓ 1
A1 (4, 5) (ECF from (a))	✓ 1
Part (c)
M1 gradient of \(AB=\frac{2-1}{6-1}\)	✓ 1
M1 gradient of \(DC=\frac{6-5}{9-4}\)	✓ 1
A1 both equal \(\frac{1}{5}\), so the lines are parallel	✓ 1
Part (d)
A1 \(\sqrt{3^2+4^2}=5\) cm	✓ 1

Common error: Stating that lines are parallel without showing the two gradients are equal.

+ Add to test✓ In your test106IGMC3-3-11r44-14Gradient, midpoint and length of a line segment Paper CHard10 marks

Calculator

A delivery van route is planned on a map grid in which 1 unit represents 1 km. A warehouse is at A(0, 8). Two shops are at B(3, −2) and C(9, 6). The van drives from A to the midpoint M of BC.

Diagram not to scale

(a)

Find the coordinates of M.

(2)

(b)

Find the equation of the line AM in the form y = mx + c.

(3)

(c)

The line AM crosses the road with equation y = 2x − 1 at the point R. Find the coordinates of R.

(3)

(d)

Calculate the distance AR. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (3 + 9)/2 or (−2 + 6)/2	✓ 1
A1 (6, 2)	✓ 1
Part (b)
M1 gradient (2 − 8)/(6 − 0)	✓ 1
A1 gradient −1	✓ 1
A1 y = −x + 8	✓ 1
Part (c)
M1 equates −x + 8 = 2x − 1	✓ 1
A1 x = 3	✓ 1
A1 y = 5, so R(3, 5)	✓ 1
Part (d)
M1 √(3² + 3²) or √18	✓ 1
A1 4.24 km	✓ 1

Common error: For intersection points, solve the two equations together and find both coordinates. Stopping after the x-value loses the final marks.

+ Add to test✓ In your test107IGMC-043-01-03Equation of a straight line, parallel and perpendicular lines Paper EEasy4 marks

No calculator

Line L has equation \(y = 4x + 1\).
Line M is parallel to line L and passes through the point (0, −3).

(a)

Write down the gradient of line M.

(1)

(b)

Write down the equation of line M.

(1)

(c)

Find the x-coordinate of the point where line M crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 4	✓ 1
Part (b)
B1: y = 4x − 3	✓ 1
Part (c)
M1: 4x − 3 = 0	✓ 1
A1: x = \(\frac{3}{4}\) or 0.75	✓ 1

Exam tip: Parallel lines have equal gradients. Only the constant term changes.

+ Add to test✓ In your test108IGMC-043-01-05Equation of a straight line, parallel and perpendicular lines Paper EEasy3 marks

No calculator

Line L has gradient \(\frac{2}{5}\).
Line M is perpendicular to line L and passes through the point (2, 3).

(a)

Write down the gradient of line M.

(1)

(b)

Find the equation of line M. Give your answer in the form \(y = mx + c\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: \(-\frac{5}{2}\)	✓ 1
Part (b)
M1: 3 = \(-\frac{5}{2}\) × 2 + c, or y − 3 = \(-\frac{5}{2}\)(x − 2)	✓ 1
A1: y = \(-\frac{5}{2}\)x + 8	✓ 1

Common error: A perpendicular gradient is the negative reciprocal. Flipping the fraction to \(\frac{5}{2}\) without changing the sign is not enough.

+ Add to test✓ In your test109IGMC3-3-09c-08Equation of a straight line, parallel and perpendicular lines Paper EEasy6 marks

No calculator

On a car park plan, one lane marking lies along the line y = 2x + 1. A second marking lies along the line 4x − 2y = 9.

(a)

Show that the two markings are parallel.

(2)

(b)

A third marking is parallel to y = 2x + 1 and passes through (3, −2). Find its equation.

(2)

(c)

Find the x-coordinate where the third marking crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges 4x − 2y = 9 to y = 2x − 4.5	✓ 1
A1 both gradients are 2, so the lines are parallel	✓ 1
Part (b)
B1 gradient 2	✓ 1
A1 y = 2x − 8 (from −2 = 6 + c)	✓ 1
Part (c)
M1 0 = 2x − 8	✓ 1
A1 x = 4	✓ 1

Exam tip: To compare gradients, write both equations in the form y = mx + c first.

+ Add to test✓ In your test110IGMC-043-00-06Gradient, midpoint and length of a line segment Paper EMedium6 marks

No calculator

The point A has coordinates (k, 3) and the point B has coordinates (5, 11). The gradient of AB is 2.

(a)

Find the value of k.

(3)

(b)

Find the length of AB. Give your answer as a surd in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{11-3}{5-k}=2\)	✓ 1
M1 for 8 = 2(5 − k) or 5 − k = 4	✓ 1
A1 k = 1	✓ 1
Part (b)
M1 for \(\sqrt{4^2+8^2}\)	✓ 1
A1 \(\sqrt{80}\)	✓ 1
A1 \(4\sqrt{5}\)	✓ 1

Exam tip: To simplify a surd, look for the largest square factor: \(80 = 16 \times 5\).

+ Add to test✓ In your test111IGMC-043-00-08Gradient, midpoint and length of a line segment Paper EMedium6 marks

No calculator

The point P has coordinates (3, −2) and the point Q has coordinates (t, 6). The length of PQ is 10 units.

(a)

Find the two possible values of t.

(4)

(b)

Given that t is positive, find the gradient of PQ.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \((t-3)^2+8^2=10^2\)	✓ 1
M1 for \((t-3)^2=36\)	✓ 1
M1 for \(t-3=\pm 6\)	✓ 1
A1 t = 9 and t = −3	✓ 1
Part (b)
M1 for \(\frac{6-(-2)}{9-3}\)	✓ 1
A1 \(\frac{4}{3}\)	✓ 1

Common error: Taking only the positive square root loses one answer. When \((t-3)^2 = 36\), \(t-3\) can be 6 or −6.

+ Add to test✓ In your test112IGMC-043-00-11Gradient, midpoint and length of a line segment Paper EMedium6 marks

No calculator

The points A(−1, −4), B(2, 2) and C(6, 10) are given.

(a)

Show that A, B and C lie on a straight line.

(3)

(b)

Find the ratio AB : BC in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for gradient AB = \(\frac{2-(-4)}{2-(-1)}=\frac{6}{3}\) = 2	✓ 1
M1 for gradient BC = \(\frac{10-2}{6-2}=\frac{8}{4}\) = 2	✓ 1
A1 Gradients are equal and B is a common point, so A, B and C are collinear	✓ 1
(Alternative: find the equation of line AB, \(y=2x-2\), and show that C satisfies it: \(2(6)-2=10\), with the marks awarded equivalently.)	—
Part (b)
M1 for AB = \(\sqrt{3^2+6^2}=\sqrt{45}=3\sqrt{5}\)	✓ 1
M1 for BC = \(\sqrt{4^2+8^2}=\sqrt{80}=4\sqrt{5}\)	✓ 1
A1 AB : BC = 3 : 4	✓ 1
(Alternative: because the points are collinear, the ratio of the x-differences 3 : 4 (or the y-differences 6 : 8) may be used. Award M1 for each correct difference and A1 for 3 : 4.)	—

Exam tip: Equal gradients alone only show that the lines are parallel. State that the two segments share point B to complete the proof.

+ Add to test✓ In your test113IGMC-043-01-06Equation of a straight line, parallel and perpendicular lines Paper EMedium8 marks

No calculator

The points A and B have coordinates A(−3, 4) and B(5, −2).

Diagram not to scale

(a)

Find the coordinates of the midpoint, M, of AB.

(2)

(b)

Find the gradient of AB.

(2)

(c)

Find the equation of the perpendicular bisector of AB. Give your answer in the form \(y = mx + c\).

(3)

(d)

Find the x-coordinate of the point where the perpendicular bisector crosses the x-axis.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: x-coordinate \(\frac{-3+5}{2} = 1\)	✓ 1
B1: y-coordinate \(\frac{4+(-2)}{2} = 1\), so M = (1, 1)	✓ 1
Part (b)
M1: \(\frac{-2-4}{5-(-3)}\) or equivalent	✓ 1
A1: \(-\frac{3}{4}\)	✓ 1
Part (c)
M1: perpendicular gradient \(\frac{4}{3}\), or the negative reciprocal of their gradient from (b)	✓ 1
M1: uses their midpoint with their perpendicular gradient, e.g. \(y - 1 = \frac{4}{3}(x - 1)\) or \(1 = \frac{4}{3} \times 1 + c\)	✓ 1
A1: \(y = \frac{4}{3}x - \frac{1}{3}\)	✓ 1
Part (d)
B1: \(\frac{1}{4}\), or follow through by correctly solving \(0 = mx + c\) for their equation in (c)	✓ 1

Exam tip: A perpendicular bisector needs two things: the negative reciprocal gradient and the midpoint. It does not pass through A or B.

+ Add to test✓ In your test114IGMC-043-01-08Equation of a straight line, parallel and perpendicular lines Paper EMedium4 marks

No calculator

Line L₁ has equation \(3x - 2y = 7\).
Line L₂ has equation \(y = kx + 5\).

(a)

Find the value of k when L₂ is parallel to L₁.

(2)

(b)

Write down the value of k when L₂ is perpendicular to L₁.

(1)

(c)

The point P on L₁ has x-coordinate 5. Find the y-coordinate of P.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: rearranges to y = \(\frac{3}{2}\)x − \(\frac{7}{2}\)	✓ 1
A1: k = \(\frac{3}{2}\)	✓ 1
Part (b)
B1: k = \(-\frac{2}{3}\)	✓ 1
Part (c)
B1: 4	✓ 1

Common error: From \(3x - 2y = 7\) the gradient is \(+\frac{3}{2}\), not \(-\frac{3}{2}\). Dividing through by −2 changes the sign of every term.

+ Add to test✓ In your test115IGMC-043-01-09Equation of a straight line, parallel and perpendicular lines Paper EMedium7 marks

No calculator

Line P passes through the points (1, 2) and (4, 8).
Line Q has equation \(2y + x = 10\).

(a)

Show that lines P and Q are perpendicular.

(3)

(b)

Find the equation of line P.

(2)

(c)

Find the coordinates of the point where lines P and Q intersect.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Gradient of P = \(\frac{8-2}{4-1}\) = 2	✓ 1
Rearranges Q to \(y = -\frac{1}{2}x + 5\), so the gradient of Q is \(-\frac{1}{2}\)	✓ 1
2 × \(-\frac{1}{2}\) = −1, so the lines are perpendicular	✓ 1
Part (b)
Uses gradient 2 with a point on P, e.g. 2 = 2 × 1 + c, 8 = 2 × 4 + c or y − 2 = 2(x − 1)	✓ 1
y = 2x	✓ 1
Part (c)
Substitutes y = 2x into Q (or another correct elimination), e.g. 2(2x) + x = 10, giving 5x = 10	✓ 1
(2, 4)	✓ 1

Exam tip: In a 'show that' question, state the conclusion clearly. Write that the product of the gradients is −1, so the lines are perpendicular.

+ Add to test✓ In your test116IGMC-043-01-12Gradient, midpoint and length of a line segment Paper EMedium5 marks

No calculator

A quadrilateral has vertices A(1, 2), B(5, 4), C(6, 8) and D(2, 6).

Diagram not to scale

(a)

Show that AB is parallel to DC.

(2)

(b)

Find the gradient of BC.

(1)

(c)

Determine whether ABCD is a rectangle. Give a reason for your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: gradient AB = \(\frac{4-2}{5-1}\) = \(\frac{1}{2}\)	✓ 1
A1: gradient DC = \(\frac{8-6}{6-2}\) = \(\frac{1}{2}\); the gradients are equal, so AB is parallel to DC	✓ 1
Part (b)
B1: \(\frac{8-4}{6-5}\) = 4	✓ 1
Part (c)
M1: a valid test for a right angle, e.g. gradient AB × gradient BC = \(\frac{1}{2}\) × 4 = 2, or stating that the perpendicular gradient to AB is −2 and comparing it with 4. Alternatively, a valid diagonal test: AC = \(\sqrt{61}\) and BD = \(\sqrt{13}\)	✓ 1
A1: a correct conclusion with its reason, e.g. 2 ≠ −1 so AB is not perpendicular to BC, or the diagonals are not equal; therefore ABCD is not a rectangle	✓ 1

Exam tip: To test for a right angle, multiply the two gradients. A product of −1 confirms that the lines are perpendicular; any other product means they are not.

+ Add to test✓ In your test117IGMC3-3-09c-03Equation of a straight line, parallel and perpendicular lines Paper EMedium6 marks

No calculator

On a map of a hillside, one zip wire W₁ lies along the line with equation x + 2y = 10. A second zip wire W₂ is to be built parallel to W₁ and passing through the point (6, −1).

(a)

Find the gradient of W₁.

(2)

(b)

Find the equation of W₂ in the form y = mx + c.

(2)

(c)

Find the x-coordinate of the point where W₂ meets the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges to y = ... , e.g. 2y = 10 − x	✓ 1
A1 gradient = −1/2	✓ 1
Part (b)
B1 gradient of W₂ is −1/2 (parallel)	✓ 1
A1 y = −¹/₂x + 2 (from −1 = −3 + c)	✓ 1
Part (c)
M1 0 = −¹/₂x + 2	✓ 1
A1 x = 4	✓ 1

Exam tip: Parallel lines have equal gradients, so you can write the gradient of the new line down immediately.

+ Add to test✓ In your test118IGMC3-3-09c-04Equation of a straight line, parallel and perpendicular lines Paper EMedium8 marks

Calculator

Two weather sensors are placed at A(1, 2) and B(5, 4) on a grid where one unit is 1 km. A relay mast M must be the same distance from A and from B, so it lies on the perpendicular bisector of AB.

Diagram not to scale

(a)

Find the coordinates of the midpoint of AB.

(2)

(b)

Find the gradient of the perpendicular bisector of AB.

(2)

(c)

Find the equation of the perpendicular bisector.

(2)

(d)

The mast stands on the line y = 0. Find its x-coordinate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x = 3	✓ 1
B1 y = 3, so (3, 3)	✓ 1
Part (b)
M1 gradient of AB = (4 − 2)/(5 − 1) = 1/2	✓ 1
A1 perpendicular gradient = −2	✓ 1
Part (c)
M1 y − 3 = −2(x − 3), using the midpoint and gradient from (a) and (b)	✓ 1
A1 y = −2x + 9	✓ 1
Part (d)
M1 0 = −2x + 9	✓ 1
A1 x = 4.5	✓ 1

Common error: The perpendicular bisector passes through the midpoint, not through A or B. Using an endpoint gives a parallel line in the wrong place.

+ Add to test✓ In your test119IGMC3-3-09r5-09Equation of a straight line, parallel and perpendicular lines Paper EMedium8 marks

No calculator

A fence for a bee corridor lies along the line L with equation 2y + x = 10. A second line M is perpendicular to L and passes through the point (3, 4).

(a)

Find the gradient of L.

(2)

(b)

Find the equation of M in the form y = mx + c.

(3)

(c)

Find the coordinates of the point where L and M meet.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearrange to y = −x/2 + 5	✓ 1
A1 −1/2	✓ 1
Part (b)
B1 gradient of M is 2	✓ 1
M1 y − 4 = 2(x − 3)	✓ 1
A1 y = 2x − 2	✓ 1
Part (c)
M1 2(2x − 2) + x = 10	✓ 1
A1 x = 2.8	✓ 1
A1 y = 3.6	✓ 1

Common error: Perpendicular gradients multiply to −1: flip the fraction and change the sign.

+ Add to test✓ In your test120IGMC3-3-09r5-10Equation of a straight line, parallel and perpendicular lines Paper EMedium6 marks

No calculator

On a Formula 1 circuit plan, two garages are at A(1, 2) and B(7, 6).

Diagram not to scale

(a)

Find the midpoint of AB.

(2)

(b)

Find the gradient of AB.

(1)

(c)

Find the equation of the perpendicular bisector of AB, giving your answer in the form ax + by = c where a, b and c are integers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x = 4	✓ 1
B1 y = 4	✓ 1
Part (b)
B1 4/6 = 2/3	✓ 1
Part (c)
B1 perpendicular gradient −3/2 (ECF from (b))	✓ 1
M1 y − 4 = −(3/2)(x − 4) (ECF from (a))	✓ 1
A1 3x + 2y = 20	✓ 1

Exam tip: A perpendicular bisector needs two ingredients: the midpoint and the negative reciprocal gradient.

+ Add to test✓ In your test121IGMC3-3-09r5-11Gradient, midpoint and length of a line segment Paper EMedium6 marks

Calculator

On a plan of a hospital site, each unit is 1 km. Three buildings are at A(1, 2), B(5, 0) and C(7, 4).

Diagram not to scale

(a)

Show that AB is perpendicular to BC.

(3)

(b)

Calculate the area of triangle ABC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient of AB = (0 − 2)/(5 − 1) = −1/2	✓ 1
A1 gradient of BC = (4 − 0)/(7 − 5) = 2	✓ 1
R1 product of gradients is −1, so the lines are perpendicular	✓ 1
Part (b)
B1 AB = √20	✓ 1
B1 BC = √20	✓ 1
A1 area = ½ × √20 × √20 = 10 km²	✓ 1

Exam tip: Show the product of the gradients explicitly; stating 'they are perpendicular' without the product earns no reasoning mark.

+ Add to test✓ In your test122IGMC3-3-09r8-09Equation of a straight line, parallel and perpendicular lines Paper EMedium6 marks

Calculator

A straight pier edge in a harbour plan lies on the line 3x + 2y = 12, with units in metres.

(a)

Find the gradient of the pier edge.

(2)

(b)

A parallel line passes through (4, −1). Find its equation in the form y = mx + c.

(2)

(c)

Find the x-coordinate where this parallel line crosses the x-axis. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges to y = −1.5x + 6	✓ 1
A1 −3/2	✓ 1
Part (b)
M1 uses gradient −1.5 with the point (4, −1)	✓ 1
A1 y = −1.5x + 5	✓ 1
Part (c)
M1 0 = −1.5x + 5	✓ 1
A1 x = 10/3 = 3.33	✓ 1

Exam tip: Parallel lines share a gradient, so find that first, then fix the intercept with the point.

+ Add to test✓ In your test123IGMC3-3-09r8-10Equation of a straight line, parallel and perpendicular lines Paper EMedium7 marks

No calculator

A wall panel in an architect's drawing is supported by a strut. A(2, 1) and B(6, 3) are the ends of the panel's top edge, with units in metres.

Diagram not to scale

(a)

Find the midpoint of AB and the gradient of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AB, in the form y = mx + c.

(3)

(c)

The strut lies along the perpendicular bisector. Find the x-coordinate where it meets the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 midpoint (4, 2)	✓ 1
B1 gradient = 2/4 = ½	✓ 1
Part (b)
M1 perpendicular gradient = −2	✓ 1
M1 y − 2 = −2(x − 4)	✓ 1
A1 y = −2x + 10	✓ 1
Part (c)
M1 0 = −2x + 10	✓ 1
A1 x = 5	✓ 1

Common error: Using the negative of the gradient rather than the negative reciprocal for perpendicular lines.

+ Add to test✓ In your test124IGMC3-3-09r12-06Equation of a straight line, parallel and perpendicular lines Paper EMedium8 marks

No calculator

In a building plan, a wall lies along the line y = 2x − 3. A support beam runs along the line through P(4, 0) that is perpendicular to the wall. Each unit represents 1 m.

(a)

Write down the gradient of the beam.

(1)

(b)

Find the equation of the beam in the form y = mx + c.

(2)

(c)

The beam meets the wall at Q. Find the coordinates of Q.

(3)

(d)

Find the exact length of PQ.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(-\frac{1}{2}\)	✓ 1
Part (b)
M1 \(y-0=-\frac{1}{2}(x-4)\)	✓ 1
A1 \(y=-\frac{1}{2}x+2\)	✓ 1
Part (c)
M1 equate \(2x-3=-\frac{1}{2}x+2\)	✓ 1
A1 x = 2	✓ 1
A1 y = 1, so Q is (2, 1)	✓ 1
Part (d)
M1 \(\sqrt{(4-2)^2+(0-1)^2}\)	✓ 1
A1 \(\sqrt{5}\) m	✓ 1

Common error: The gradient of a perpendicular line is the negative reciprocal: flip the fraction and change the sign.

+ Add to test✓ In your test125IGMC3-3-09r12-08Equation of a straight line, parallel and perpendicular lines Paper EMedium6 marks

Calculator

A boardwalk edge in a marine park lies along the line L with equation 3x + 4y = 24. Each unit on the grid represents 1 m.

(a)

Find the gradient of L.

(2)

(b)

A second edge is parallel to L and passes through (8, −5). Find its equation in the form y = mx + c.

(2)

(c)

This second edge crosses the x-axis. Find the x-coordinate of the crossing point, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearrange to \(y=-\frac{3}{4}x+6\)	✓ 1
A1 \(-\frac{3}{4}\)	✓ 1
Part (b)
M1 use gradient \(-\frac{3}{4}\) with the point, e.g. \(-5=-\frac{3}{4}(8)+c\)	✓ 1
A1 \(y=-\frac{3}{4}x+1\)	✓ 1
Part (c)
M1 substitute y = 0: \(0=-\frac{3}{4}x+1\)	✓ 1
A1 1.33	✓ 1

Exam tip: Parallel lines have equal gradients, so reuse the gradient you found in the first part.

+ Add to test✓ In your test126IGMC3-3-09r12-09Equation of a straight line, parallel and perpendicular lines Paper EMedium6 marks

Calculator

Two sensors in a coral survey are at S(−2, 5) and T(6, 1). Each unit on the grid represents 1 m. A third sensor must be placed so that it is the same distance from S and T.

Diagram not to scale

(a)

Find the midpoint of ST.

(1)

(b)

Find the gradient of ST.

(1)

(c)

Find the equation of the line of all points equidistant from S and T, in the form y = mx + c.

(2)

(d)

The third sensor is placed on this line at the point where y = 0. Find its x-coordinate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 (2, 3)	✓ 1
Part (b)
B1 \(-\frac{1}{2}\)	✓ 1
Part (c)
M1 perpendicular gradient 2 through (2, 3)	✓ 1
A1 y = 2x − 1	✓ 1
Part (d)
M1 0 = 2x − 1	✓ 1
A1 0.5	✓ 1

Common error: Points equidistant from two points lie on the perpendicular bisector, not on the line joining them.

+ Add to test✓ In your test127IGMC3-3-10r15-09Equation of a straight line, parallel and perpendicular lines Paper EMedium7 marks

No calculator

In a game, a wall runs from A(2, 1) to B(6, 9). A line of sight is the perpendicular bisector of AB.

Diagram not to scale

(a)

Find the midpoint of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AB in the form y = mx + c.

(3)

(c)

Find the x-coordinate of the point where the line of sight meets the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ((2 + 6)/2, (1 + 9)/2)	✓ 1
A1 (4, 5)	✓ 1
Part (b)
M1 gradient of AB = (9 − 1)/(6 − 2) = 2	✓ 1
M1 perpendicular gradient −½ and use of point (4, 5): y − 5 = −½(x − 4)	✓ 1
A1 y = −½x + 7	✓ 1
Part (c)
M1 0 = −½x + 7	✓ 1
A1 x = 14	✓ 1

Exam tip: The gradients of perpendicular lines multiply to give −1, so the perpendicular gradient is the negative reciprocal.

+ Add to test✓ In your test128IGMC3-3-10r15-11Equation of a straight line, parallel and perpendicular lines Paper EMedium9 marks

No calculator

A water-supply engineer models a straight pipe L by the equation 2y − 6x = 10.

(a)

Write down the gradient and the y-intercept of L.

(2)

(b)

Find the equation of the line parallel to L that passes through (2, −1).

(2)

(c)

Show that the line 3y + x = 7 is perpendicular to L.

(2)

(d)

Find the coordinates of the point where L meets the line 3y + x = 7.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearrange to y = 3x + 5	✓ 1
A1 gradient 3 and y-intercept 5	✓ 1
Part (b)
M1 −1 = 3 × 2 + c	✓ 1
A1 y = 3x − 7	✓ 1
Part (c)
M1 rearrange to y = −⅓x + 7/3, so gradient −1/3	✓ 1
A1 3 × (−1/3) = −1, so the lines are perpendicular	✓ 1
Part (d)
M1 substitute y = 3x + 5 into 3y + x = 7	✓ 1
A1 10x = −8, so x = −0.8	✓ 1
A1 y = 2.6	✓ 1

Common error: Dividing every term by 2 when rearranging is essential: 2y = 6x + 10 gives y = 3x + 5, not y = 6x + 10.

+ Add to test✓ In your test129IGMC3-3-10r21-06Equation of a straight line, parallel and perpendicular lines Paper EMedium6 marks

No calculator

On a telecoms planning map, a cable runs along the line through P(1, 3) and Q(5, 11). A second cable passes through R(6, 1) and is perpendicular to the cable through P and Q.

(a)

Find the gradient of PQ.

(2)

(b)

Find the equation of the second cable, giving your answer in the form \(y=mx+c\).

(3)

(c)

Find the x-coordinate of the point where the second cable crosses the x-axis.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{11-3}{5-1}\)	✓ 1
A1 2	✓ 1
Part (b)
B1 perpendicular gradient \(-\frac{1}{2}\)	✓ 1
M1 \(y-1=-\frac{1}{2}(x-6)\)	✓ 1
A1 \(y=-\frac{1}{2}x+4\)	✓ 1
Part (c)
A1 8	✓ 1

Exam tip: Perpendicular gradients multiply to give −1. Flip the fraction and change the sign.

+ Add to test✓ In your test130IGMC3-3-10r21-08Equation of a straight line, parallel and perpendicular lines Paper EMedium5 marks

No calculator

A food technologist buys x kg of flour at €2 per kg and y kg of sugar at €4 per kg. The total cost is exactly €40, so \(2x+4y=40\).

(a)

Rearrange the equation into the form \(y=mx+c\).

(2)

(b)

State the gradient of the line and explain what it means in this context.

(2)

(c)

Find the mass of sugar bought when 14 kg of flour is bought.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(4y=40-2x\)	✓ 1
A1 \(y=-\frac{1}{2}x+10\)	✓ 1
Part (b)
B1 gradient \(-\frac{1}{2}\)	✓ 1
B1 each extra 1 kg of flour bought means 0.5 kg less sugar can be bought	✓ 1
Part (c)
A1 \(y=-7+10=3\) kg	✓ 1

Exam tip: Interpreting a gradient: say 'for each 1 unit increase in x, y changes by ...' and use the units from the question.

+ Add to test✓ In your test131IGMC3-3-10r21-09Equation of a straight line, parallel and perpendicular lines Paper EMedium5 marks

No calculator

On a map of a mountain ridge, one ridgeline has equation \(3y-6x=9\). A second ridgeline is parallel to the first and passes through the point (4, −1).

(a)

Find the gradient of the first ridgeline.

(2)

(b)

Find the equation of the second ridgeline, in the form \(y=mx+c\).

(2)

(c)

Show that the point (5, 1) lies on the second ridgeline.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(y=2x+3\)	✓ 1
A1 2	✓ 1
Part (b)
M1 \(-1=2\times4+c\)	✓ 1
A1 \(y=2x-9\)	✓ 1
Part (c)
A1 \(2\times5-9=1\)	✓ 1

Common error: Using the coefficient of x as the gradient before rearranging: in \(3y-6x=9\) the gradient is \(6\div3=2\), not 6.

+ Add to test✓ In your test132IGMC3-3-11r29-06Equation of a straight line, parallel and perpendicular lines Paper EMedium6 marks

No calculator

In a computer game, a wall lies along the line x + 2y = 6. A laser beam is fired from the point (4, 3) and travels parallel to the wall.

(a)

Find the gradient of the wall.

(2)

(b)

Find the equation of the laser beam in the form y = mx + c.

(2)

(c)

Find the coordinates of the point where the beam crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2y = −x + 6	✓ 1
A1 −1/2	✓ 1
Part (b)
M1 y − 3 = −½(x − 4) (parallel lines have equal gradients)	✓ 1
A1 y = −½x + 5	✓ 1
Part (c)
M1 substitute y = 0	✓ 1
A1 (10, 0)	✓ 1

Common error: Parallel lines have the same gradient; do not use the negative reciprocal, which is for perpendicular lines.

+ Add to test✓ In your test133IGMC3-3-11r29-07Equation of a straight line, parallel and perpendicular lines Paper EMedium10 marks

Calculator

Two water pumping stations are at A(2, 1) and B(8, 5) on a map grid where 1 unit represents 1 km. A new storage tower T must be the same distance from A and from B.

Diagram not to scale

(a)

Find the gradient of AB.

(2)

(b)

Find the midpoint of AB.

(1)

(c)

Find the equation of the perpendicular bisector of AB, giving your answer in the form ax + by = c, where a, b and c are integers.

(3)

(d)

The tower is built at T(3, 6). Show that T lies on the perpendicular bisector.

(2)

(e)

Calculate the distance TA.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (5 − 1)/(8 − 2)	✓ 1
A1 2/3	✓ 1
Part (b)
B1 (5, 3)	✓ 1
Part (c)
M1 perpendicular gradient = −3/2	✓ 1
M1 y − 3 = −3/2 (x − 5)	✓ 1
A1 3x + 2y = 21	✓ 1
Part (d)
M1 substitute: 3(3) + 2(6)	✓ 1
A1 = 21, so T lies on the line	✓ 1
Part (e)
M1 √(1² + 5²)	✓ 1
A1 5.10 km (3 s.f.)	✓ 1

Exam tip: Every point on the perpendicular bisector of AB is equally far from A and B, so you can check your equation using a point you know is on it, such as the midpoint.

+ Add to test✓ In your test134IGMC3-3-11r29-09Equation of a straight line, parallel and perpendicular lines Paper EMedium8 marks

Calculator

In a game, a ray of light passes through the point A(−1, 3) and is perpendicular to the mirror line m, whose equation is 3y = x + 6.

Diagram not to scale

(a)

Find the equation of the ray.

(3)

(b)

Find the coordinates of the point where the ray meets the mirror.

(3)

(c)

Calculate the distance from A to the point where the ray meets the mirror.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient of m is 1/3, so the ray has gradient −3	✓ 1
M1 y − 3 = −3(x + 1)	✓ 1
A1 y = −3x	✓ 1
Part (b)
M1 substitute: 3(−3x) = x + 6	✓ 1
A1 x = −0.6	✓ 1
A1 y = 1.8	✓ 1
Part (c)
M1 √(0.4² + 1.2²)	✓ 1
A1 1.26 units (3 s.f.)	✓ 1

Common error: Take the reciprocal and change the sign to get a perpendicular gradient; multiplying the two gradients should give −1.

+ Add to test✓ In your test135IGMC3-3-11r35-08Equation of a straight line, parallel and perpendicular lines Paper EMedium8 marks

No calculator

A straight footpath on a vineyard plan joins gate A(1, 2) to gate B(7, 6).

Diagram not to scale

(a)

Find the gradient of AB.

(2)

(b)

A second path is perpendicular to AB and passes through B. Write down its gradient.

(1)

(c)

Find the equation of the second path. Give your answer in the form ax + by = c, where a, b and c are integers.

(3)

(d)

The second path crosses the x-axis at C. Find the coordinates of C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (6 − 2) ÷ (7 − 1)	✓ 1
A1 2/3	✓ 1
Part (b)
B1 −3/2 (ECF from (a))	✓ 1
Part (c)
M1 y − 6 = −3/2(x − 7), or y = −3/2x + c with substitution of (7, 6)	✓ 1
M1 multiplies through to remove fractions	✓ 1
A1 3x + 2y = 33	✓ 1
Part (d)
M1 substitutes y = 0 giving 3x = 33 (ECF from (c))	✓ 1
A1 (11, 0)	✓ 1

Exam tip: Perpendicular gradients multiply to give −1: turn the fraction upside down and change the sign.

+ Add to test✓ In your test136IGMC3-3-11r35-09Equation of a straight line, parallel and perpendicular lines Paper EMedium6 marks

No calculator

On a plan of a farm, a fence L₁ lies along the line 3x − 2y = 12.

(a)

Find the gradient of L₁.

(2)

(b)

A second fence L₂ is parallel to L₁ and passes through (4, −1). Find the equation of L₂ in the form y = mx + c.

(2)

(c)

The point (a, 8) lies on L₂. Find the value of a.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges to y = 3/2 x − 6	✓ 1
A1 3/2	✓ 1
Part (b)
M1 y = 3/2 x + c and −1 = 6 + c (ECF from (a))	✓ 1
A1 y = 3/2 x − 7	✓ 1
Part (c)
M1 8 = 3/2 a − 7 (ECF from (b))	✓ 1
A1 a = 10	✓ 1

Common error: The gradient of 3x − 2y = 12 is not 3. Rearrange to make y the subject before reading off the gradient.

+ Add to test✓ In your test137IGMC3-3-10r43-06Equation of a straight line, parallel and perpendicular lines Paper EMedium7 marks

No calculator

A tailor marks a seam AB on a pattern grid, where A is (2, 1) and B is (6, 9). The point C is (5, 4).

Diagram not to scale

(a)

Find the gradient of AB.

(2)

(b)

Find the equation of the line through C that is parallel to AB. Give your answer in the form y = mx + c.

(2)

(c)

Find the equation of the line through B that is perpendicular to AB. Give your answer in the form y = mx + c.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (9 − 1) / (6 − 2)	✓ 1
A1 2	✓ 1
Part (b)
M1 y − 4 = 2(x − 5) (parallel lines have the same gradient)	✓ 1
A1 y = 2x − 6	✓ 1
Part (c)
B1 gradient = −1/2	✓ 1
M1 y − 9 = −(1/2)(x − 6)	✓ 1
A1 y = −(1/2)x + 12	✓ 1

Common error: A perpendicular gradient is the negative reciprocal: turn the fraction upside down and change the sign.

+ Add to test✓ In your test138IGMC3-3-10r43-12Equation of a straight line, parallel and perpendicular lines Paper EMedium6 marks

No calculator

Two lines of stitching on a design are given by L₁: kx + 2y = 9 and L₂: 3x − y = 4, where k is a constant.

(a)

Find the gradient of L₂.

(1)

(b)

Find the value of k for which L₁ is parallel to L₂.

(2)

(c)

Find the value of k for which L₁ is perpendicular to L₂.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3 (from y = 3x − 4)	✓ 1
Part (b)
M1 gradient of L₁ = −k/2 and −k/2 = 3	✓ 1
A1 k = −6	✓ 1
Part (c)
M1 product of gradients = −1, so (−k/2) × 3 = −1	✓ 1
A1 −3k/2 = −1	✓ 1
A1 k = 2/3 (≈ 0.667)	✓ 1

Common error: For perpendicular lines the product of the gradients is −1; for parallel lines the gradients are equal.

+ Add to test✓ In your test139IGMC3-3-11r44-09Equation of a straight line, parallel and perpendicular lines Paper EMedium7 marks

No calculator

In a supermarket plan, one aisle runs along the straight line through (2, 1) and (6, 9).

(a)

Find the gradient of this aisle.

(2)

(b)

A second aisle is parallel to the first and passes through (1, −5). Find its equation in the form y = mx + c.

(3)

(c)

The point (k, 15) lies on the second aisle. Find k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (9 − 1)/(6 − 2)	✓ 1
A1 2	✓ 1
Part (b)
B1 gradient 2 (parallel lines have equal gradients)	✓ 1
M1 substitutes (1, −5), e.g. −5 = 2 × 1 + c	✓ 1
A1 y = 2x − 7	✓ 1
Part (c)
M1 15 = 2k − 7	✓ 1
A1 k = 11	✓ 1

Common error: Parallel lines have the same gradient. Do not use the negative reciprocal, which is only for perpendicular lines.

+ Add to test✓ In your test140IGMC3-3-11r44-11Gradient, midpoint and length of a line segment Paper EMedium6 marks

Calculator

Three vaccination points are at A(1, 1), B(5, 3) and C(3, 7) on a map grid in which 1 unit represents 1 km.

Diagram not to scale

(a)

Show that angle ABC is a right angle.

(3)

(b)

Calculate the area of triangle ABC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient of AB = 2/4 = 1/2	✓ 1
M1 gradient of BC = 4/(−2) = −2	✓ 1
R1 product of gradients = −1, so AB and BC are perpendicular	✓ 1
Part (b)
M1 attempt at length AB or BC, e.g. √20	✓ 1
M1 area = ½ × AB × BC	✓ 1
A1 10 km²	✓ 1

Common error: Showing a right angle needs a stated conclusion: the product of the gradients is −1, so the lines are perpendicular. Equal lengths alone do not prove it.

+ Add to test✓ In your test141IGMC3-3-11r44-12Equation of a straight line, parallel and perpendicular lines Paper EMedium8 marks

No calculator

Two wind turbines stand at A(−2, 6) and B(4, 0) on a map where 1 unit represents 1 km. A maintenance track passes through the midpoint M of AB and is parallel to a road with equation 4x + 2y = 7.

Diagram not to scale

(a)

Find the coordinates of M.

(2)

(b)

Find the gradient of the road.

(2)

(c)

Find the equation of the track in the form y = mx + c.

(2)

(d)

Find the x-coordinate of the point where the track crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (−2 + 4)/2 or (6 + 0)/2	✓ 1
A1 (1, 3)	✓ 1
Part (b)
M1 rearranges to y = −2x + 3.5	✓ 1
A1 −2	✓ 1
Part (c)
M1 gradient −2 and point (1, 3), e.g. y − 3 = −2(x − 1)	✓ 1
A1 y = −2x + 5	✓ 1
Part (d)
M1 sets y = 0, so −2x + 5 = 0	✓ 1
A1 x = 2.5	✓ 1

Common error: To read the gradient from 4x + 2y = 7, rearrange to make y the subject. The gradient is not 4 or 2.

+ Add to test✓ In your test142IGMC3-3-10r47-05Equation of a straight line, parallel and perpendicular lines Paper EMedium10 marks

No calculator

A footpath follows the line x + 2y = 10. A surveyor stands at the point P(5, 0) and wants to walk to the path by the shortest route. Units are kilometres.

Diagram not to scale

(a)

Find the gradient of the path.

(2)

(b)

Find the equation of the line through P that is perpendicular to the path.

(3)

(c)

Find the coordinates of the point where this line meets the path.

(3)

(d)

Find the shortest distance from P to the path. Give your answer as a surd.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearrange to y = −x/2 + 5	✓ 1
A1 −1/2	✓ 1
Part (b)
B1 perpendicular gradient 2 (ECF from (a))	✓ 1
M1 y − 0 = 2(x − 5)	✓ 1
A1 y = 2x − 10	✓ 1
Part (c)
M1 x + 2(2x − 10) = 10	✓ 1
A1 x = 6	✓ 1
A1 y = 2, so (6, 2)	✓ 1
Part (d)
M1 √((6 − 5)² + (2 − 0)²) (ECF from (c))	✓ 1
A1 √5 km (= 2.236 km)	✓ 1

Exam tip: The shortest route from a point to a line is along the perpendicular, so the answer to (d) uses the perpendicular line from (b).

+ Add to test✓ In your test143IGMC3-3-10r47-08Equation of a straight line, parallel and perpendicular lines Paper EMedium10 marks

Calculator

Telecoms towers are at A(−4, 1) and B(6, 5). A new mast M must be the same distance from A as from B and must be on the x-axis. Units are kilometres.

Diagram not to scale

(a)

Find the midpoint of AB and the gradient of AB.

(3)

(b)

Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c.

(3)

(c)

Find the coordinates of M.

(2)

(d)

Calculate the distance from M to A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint ((−4 + 6)/2, (1 + 5)/2)	✓ 1
A1 (1, 3)	✓ 1
A1 gradient = 4/10 = 0.4	✓ 1
Part (b)
B1 perpendicular gradient = −1/0.4 = −2.5 (ECF from (a))	✓ 1
M1 y − 3 = −2.5(x − 1)	✓ 1
A1 y = −2.5x + 5.5	✓ 1
Part (c)
M1 substitute y = 0 to get 0 = −2.5x + 5.5	✓ 1
A1 (2.2, 0)	✓ 1
Part (d)
M1 √((2.2 + 4)² + (0 − 1)²) = √39.44	✓ 1
A1 6.28 km	✓ 1

Common error: Every point on the perpendicular bisector is equidistant from A and B; the bisector passes through the midpoint, not through A or B.

+ Add to test✓ In your test144IGMC3-3-10r47-13Equation of a straight line, parallel and perpendicular lines Paper EMedium8 marks

Calculator

On a coordinate grid centred on a base station at the origin O, two cable lines are modelled by the equations 4y = 3x + 8 and 3y + 4x = 15. Units are kilometres.

(a)

Show that the two lines are perpendicular.

(3)

(b)

Find the coordinates of the point where the two lines meet.

(3)

(c)

Calculate the distance from this point to the base station O.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient of the first line = 3/4	✓ 1
M1 gradient of the second line = −4/3	✓ 1
R1 product = −1, so the lines are perpendicular	✓ 1
Part (b)
M1 eliminate a variable, for example 9x + 24 = 60 − 16x	✓ 1
A1 x = 1.44	✓ 1
A1 y = 3.08	✓ 1
Part (c)
M1 √(1.44² + 3.08²) (ECF from (b))	✓ 1
A1 3.4 km	✓ 1

Exam tip: Check your intersection point by substituting both coordinates into the other equation as well.

+ Add to test✓ In your test145IGMC-043-00-12Gradient, midpoint and length of a line segment Paper EHard6 marks

No calculator

The point P(a, 0) lies on the x-axis. P is the same distance from A(1, 5) as it is from B(7, 3).

(a)

Find the value of a. Give your answer as a fraction.

(4)

(b)

Find the distance PA. Give your answer in the form \(\frac{p\sqrt{q}}{r}\), where p, q and r are integers.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \((a-1)^2+5^2\) or \((a-7)^2+3^2\)	✓ 1
M1 for \((a-1)^2+25=(a-7)^2+9\)	✓ 1
M1 for expanding to \(-2a+26=-14a+58\) or \(12a=32\)	✓ 1
A1 \(a=\frac{8}{3}\)	✓ 1
Part (b)
M1 for \(\sqrt{\left(\frac{5}{3}\right)^2+5^2}\) or \(\sqrt{\frac{250}{9}}\)	✓ 1
A1 \(\frac{5\sqrt{10}}{3}\)	✓ 1

Exam tip: Set the squared distances equal to avoid square roots. The \(a^2\) terms cancel, leaving a linear equation.

+ Add to test✓ In your test146IGMC-043-00-13Gradient, midpoint and length of a line segment Paper EHard10 marks

No calculator

ABCD is a parallelogram with A(1, 2), B(6, 4) and C(8, 9). The diagonals of a parallelogram bisect each other.

Diagram not to scale

(a)

Find the coordinates of D.

(3)

(b)

Show that AB = BC.

(2)

(c)

Show that the diagonals AC and BD are perpendicular.

(3)

(d)

Your results show that ABCD is a rhombus. The area of a rhombus is half the product of the lengths of its diagonals. Find the area of ABCD.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of AC = (4.5, 5.5) or translation A→B is (+5, +2)	✓ 1
\(\frac{6+x}{2}=4.5\) and \(\frac{4+y}{2}=5.5\) or D = C − (5, 2), using DC = AB	✓ 1
D(3, 7)	✓ 1
Part (b)
AB = \(\sqrt{5^2+2^2}\) or BC = \(\sqrt{2^2+5^2}\)	✓ 1
Both equal \(\sqrt{29}\), so AB = BC	✓ 1
Part (c)
Gradient AC = \(\frac{9-2}{8-1}\) = 1	✓ 1
Gradient BD = \(\frac{7-4}{3-6}\) = −1	✓ 1
1 × (−1) = −1, so the diagonals are perpendicular	✓ 1
Part (d)
AC = \(\sqrt{98}=7\sqrt{2}\) and BD = \(\sqrt{18}=3\sqrt{2}\)	✓ 1
Area = \(\frac{1}{2}\times7\sqrt{2}\times3\sqrt{2}\) = 21 square units	✓ 1

Common error: Do not take D as the fourth vertex in the wrong order, for example A + B − C. In parallelogram ABCD the vertices go round in order, so A and C are opposite. That gives D = A + C − B, which is the same as using the midpoint of AC.

+ Add to test✓ In your test147IGMC-043-00-14Gradient, midpoint and length of a line segment Paper EHard8 marks

No calculator

The line segment AB joins A(2, −1) to B(10, 5).

Diagram not to scale

(a)

Find the length of AB.

(2)

(b)

The point P lies on AB such that AP : PB = 1 : 3. Find the coordinates of P.

(3)

(c)

The line through A and B crosses the x-axis at the point R. Find the x-coordinate of R. Give your answer as a fraction.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for \(\sqrt{8^2+6^2}\)	✓ 1
A1 10	✓ 1
Part (b)
M1 for AP is \(\frac{1}{4}\) of AB	✓ 1
M1 for \(\frac{1}{4}\) of (8, 6) = (2, 1.5) added to A	✓ 1
A1 (4, 0.5)	✓ 1
Part (c)
M1 for gradient \(\frac{6}{8}=\frac{3}{4}\)	✓ 1
M1 for \(0-(-1)=\frac{3}{4}(x-2)\) or equivalent	✓ 1
A1 \(x=\frac{10}{3}\)	✓ 1

Common error: For a ratio of 1 : 3, P is \(\frac{1}{4}\) of the way along, not \(\frac{1}{3}\). Add the parts of the ratio to find the fraction.

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Calculator

The points P and Q have coordinates P(1, −1) and Q(7, 2).
Line L passes through the point R(−2, 7) and is perpendicular to PQ.

Diagram not to scale

(a)

Find the equation of line L in the form \(y = mx + c\).

(3)

(b)

Find the coordinates of the point, F, where L meets the line PQ.

(3)

(c)

Calculate the shortest distance from R to the line PQ.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: gradient PQ = \(\frac{3}{6}\) = \(\frac{1}{2}\)	✓ 1
M1: uses gradient −2 with (−2, 7): 7 = −2 × (−2) + c	✓ 1
A1: y = −2x + 3	✓ 1
Part (b)
M1: equation of PQ: y = \(\frac{1}{2}\)x − \(\frac{3}{2}\)	✓ 1
M1: −2x + 3 = \(\frac{1}{2}\)x − \(\frac{3}{2}\)	✓ 1
A1: F = (1.8, −0.6)	✓ 1
Part (c)
M1: \(\sqrt{(1.8-(-2))^2 + (-0.6-7)^2}\)	✓ 1
A1: 8.50	✓ 1

Exam tip: The shortest distance from a point to a line is measured along the perpendicular. Find the foot of the perpendicular first, then use the distance formula.

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Calculator

The line L has equation \(5x + 2y = 20\). It meets the x-axis at A and the y-axis at B.

(a)

Find the coordinates of A and of B.

(2)

(b)

Calculate the length of AB.

(2)

(c)

Find the equation of the line through the origin that is perpendicular to L.

(2)

(d)

Calculate the acute angle between L and the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: A(4, 0)	✓ 1
B1: B(0, 10)	✓ 1
Part (b)
M1: \(\sqrt{4^2 + 10^2}\) or \(\sqrt{116}\)	✓ 1
A1: 10.8 (accept 10.77… or exact \(2\sqrt{29}\))	✓ 1
Part (c)
M1: gradient of L = \(-\frac{5}{2}\), so the perpendicular gradient is \(\frac{2}{5}\)	✓ 1
A1: \(y = \frac{2}{5}x\) (accept equivalent forms, e.g. \(2x - 5y = 0\))	✓ 1
Part (d)
M1: \(\tan\theta = \frac{10}{4}\) (or \(\tan\theta = \frac{5}{2}\), from the magnitude of the gradient of L)	✓ 1
A1: 68.2°	✓ 1

Common error: Using \(\tan\theta = \frac{4}{10}\) gives 21.8°, which is the angle with the y-axis. The angle with the x-axis uses the vertical side over the horizontal side, which is the magnitude of the gradient.

+ Add to test✓ In your test150IGMC-043-01-14Equation of a straight line, parallel and perpendicular lines Paper EHard8 marks

No calculator

Line L₁ has equation \(4x - 3y = 12\).
Line L₂ is perpendicular to L₁ and passes through the point (2, −1).

(a)

Show that the equation of L₂ can be written as \(3x + 4y = 2\).

(3)

(b)

The point (k, 3) lies on L₂. Find the value of k.

(2)

(c)

Find the exact coordinates of the point where L₁ and L₂ intersect.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: gradient of L₁ = \(\frac{4}{3}\)	✓ 1
M1: y + 1 = \(-\frac{3}{4}\)(x − 2)	✓ 1
A1: 4y + 4 = −3x + 6, leading to 3x + 4y = 2	✓ 1
Part (b)
M1: 3k + 12 = 2	✓ 1
A1: k = \(-\frac{10}{3}\)	✓ 1
Part (c)
M1: eliminates one variable, e.g. 16x − 12y = 48 and 9x + 12y = 6	✓ 1
M1: 25x = 54, so x = \(\frac{54}{25}\)	✓ 1
A1: (\(\frac{54}{25}\), \(-\frac{28}{25}\))	✓ 1

Exam tip: When an exact answer is required, keep fractions throughout. Check the answer by substituting it back into both equations.

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Calculator

A straight mountain road passes through A(2, −1) and B(k, 5) on a map with units in kilometres. The gradient of the road is 3/2. A path L is built perpendicular to the road, passing through B.

(a)

Find the value of k.

(3)

(b)

Find the equation of L in the form y = mx + c.

(3)

(c)

L meets the axes at two points. Together with the origin they form a triangle. Calculate the area of this triangle.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient expression (5 − (−1))/(k − 2)	✓ 1
M1 6/(k − 2) = 3/2	✓ 1
A1 k = 6	✓ 1
Part (b)
B1 gradient of L = −2/3	✓ 1
M1 substitutes B(6, 5), e.g. 5 = −4 + c	✓ 1
A1 y = −²/₃x + 9	✓ 1
Part (c)
B1 intercepts (13.5, 0) and (0, 9)	✓ 1
M1 area = ½ × 13.5 × 9	✓ 1
A1 60.75 km² (60.8 to 3 s.f.)	✓ 1

Exam tip: For perpendicular lines the gradients multiply to −1: flip the fraction and change the sign.

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Calculator

The straight bank of a river lies along the line x + 2y = 10, with distances in kilometres. A water sample is taken at P(3, −1). A walker takes the shortest route from P to the bank, which meets the bank at right angles at the point F.

Diagram not to scale

(a)

Find the gradient of the river bank.

(2)

(b)

Find the equation of the line PF.

(3)

(c)

Find the coordinates of F.

(3)

(d)

Calculate the distance PF.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges to y = 5 − ¹/₂x	✓ 1
A1 −1/2	✓ 1
Part (b)
B1 gradient of PF = 2	✓ 1
M1 y − (−1) = 2(x − 3)	✓ 1
A1 y = 2x − 7	✓ 1
Part (c)
M1 substitutes y = 2x − 7 into x + 2y = 10, giving 5x − 14 = 10	✓ 1
A1 x = 4.8	✓ 1
A1 y = 2.6, so F = (4.8, 2.6)	✓ 1
Part (d)
M1 √((4.8 − 3)² + (2.6 + 1)²) (ECF from (c))	✓ 1
A1 4.02 km	✓ 1

Exam tip: Solve simultaneous equations by substituting the expression for y into the other line. Keep exact values like 4.8 rather than rounding.

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No calculator

On a grid of a solar farm, panels are placed at the corners of a quadrilateral with vertices A(1, 1), B(5, 3) and C(3, 7), and a fourth corner D such that ABCD is a square.

Diagram not to scale

(a)

Show that AB is perpendicular to BC.

(3)

(b)

Show that AB = BC and write the length as 2√5.

(3)

(c)

Find the coordinates of D.

(2)

(d)

Find the area of the square.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient of AB = 2/4 = 1/2	✓ 1
M1 gradient of BC = 4/(−2) = −2	✓ 1
A1 product of gradients = −1, so perpendicular	✓ 1
Part (b)
M1 AB² = 4² + 2² = 20	✓ 1
M1 BC² = (−2)² + 4² = 20	✓ 1
A1 both lengths √20 = 2√5	✓ 1
Part (c)
M1 D = A + C − B, or the same movement as from B to C applied from A	✓ 1
A1 D = (−1, 5)	✓ 1
Part (d)
M1 (2√5)² = 4² + 2²	✓ 1
A1 20 square units	✓ 1

Exam tip: Leave lengths as surds until the end; squaring √20 returns exactly 20 with no rounding.

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Calculator

On a Formula 1 circuit plan, one unit represents 10 m. Two team garages are at A(−2, 1) and B(6, 5). A refuelling point C is to be placed on the x-axis (the pit road) so that it is the same distance from A and from B.

Diagram not to scale

(a)

Find the midpoint of AB.

(2)

(b)

Find the equation of the perpendicular bisector of AB.

(4)

(c)

Find the coordinates of C.

(2)

(d)

Calculate the real distance from A to C, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x = 2	✓ 1
B1 y = 3	✓ 1
Part (b)
B1 gradient of AB = 4/8 = 1/2	✓ 1
B1 perpendicular gradient = −2	✓ 1
M1 y − 3 = −2(x − 2) (ECF from (a))	✓ 1
A1 y = −2x + 7	✓ 1
Part (c)
M1 substitute y = 0, so 2x = 7	✓ 1
A1 C(3.5, 0)	✓ 1
Part (d)
M1 √(5.5² + 1²) = 5.59… units (ECF from (c))	✓ 1
A1 55.9 m	✓ 1

Exam tip: Any point on a perpendicular bisector is equally far from both end points, so intersecting it with an axis gives a point with a given property.

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Calculator

On a hospital site plan, each unit is 1 km. A helipad, a ward, a laboratory and a car park are at A(1, 1), B(7, 3), C(8, 7) and D respectively, and ABCD is a parallelogram.

Diagram not to scale

(a)

Find the midpoint of the diagonal AC.

(2)

(b)

Use your answer to (a) to find the coordinates of D.

(2)

(c)

Show that DC is parallel to AB.

(3)

(d)

Calculate the length of the diagonal AC, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x = 4.5	✓ 1
B1 y = 4	✓ 1
Part (b)
M1 the diagonals of a parallelogram bisect each other, so midpoint of BD is (4.5, 4)	✓ 1
A1 D(2, 5)	✓ 1
Part (c)
M1 gradient of AB = 2/6	✓ 1
M1 gradient of DC = (7 − 5)/(8 − 2)	✓ 1
A1 both gradients are 1/3, so the lines are parallel	✓ 1
Part (d)
M1 √(7² + 6²)	✓ 1
A1 9.22 km	✓ 1

Common error: Parallel lines have equal gradients; check the sign and order of the subtraction in both gradients.

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Calculator

A line l₁ has equation y = 2x − 6. A second line l₂ has equation 4x + ky = 10, where k is a constant, and is parallel to l₁. The plan is used for a row of hospital beds and one unit is 1 metre.

(a)

Find the value of k.

(3)

(b)

A line q is perpendicular to l₁ and passes through the point where l₁ crosses the x-axis. Find the equation of q.

(3)

(c)

Calculate the distance between the point where q crosses the y-axis and the point where q crosses the x-axis, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearrange to y = −(4/k)x + 10/k	✓ 1
M1 −4/k = 2	✓ 1
A1 k = −2	✓ 1
Part (b)
B1 l₁ crosses the x-axis at (3, 0)	✓ 1
B1 gradient of q is −1/2	✓ 1
A1 y = −0.5x + 1.5	✓ 1
Part (c)
B1 y-intercept of q is (0, 1.5) (ECF from (b))	✓ 1
M1 √(3² + 1.5²)	✓ 1
A1 3.35 m	✓ 1

Exam tip: Writing both lines in y = mx + c form makes the parallel condition a simple comparison of gradients.

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Calculator

A straight road has equation x + 3y = 9. A depot is at P(4, −2). Units are km. A service track runs from the depot to meet the road at right angles at point F.

Diagram not to scale

(a)

Find the gradient of the road.

(2)

(b)

Find the equation of the service track.

(2)

(c)

Find the coordinates of F.

(3)

(d)

Calculate the length of the service track PF. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges to y = −x/3 + 3	✓ 1
A1 −1/3	✓ 1
Part (b)
M1 gradient 3 and point (4, −2)	✓ 1
A1 y = 3x − 14	✓ 1
Part (c)
M1 3x − 14 = −x/3 + 3 (or x + 3(3x − 14) = 9)	✓ 1
A1 x = 5.1	✓ 1
A1 y = 1.3	✓ 1
Part (d)
M1 √((5.1 − 4)² + (1.3 − (−2))²) = √12.1	✓ 1
A1 3.48 km	✓ 1

Exam tip: The shortest distance from a point to a line is along the perpendicular, so the foot of the perpendicular is the key step.

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Calculator

In a marine study, line l₁ has equation 3x − ky = 12, where k is a constant. Line l₂ passes through (2, 1) and (5, 7). The lines are perpendicular. Units are km.

(a)

Find the gradient of l₂.

(2)

(b)

Find the value of k.

(3)

(c)

Show that l₁ and l₂ intersect at (2, 1).

(2)

(d)

l₁ meets the x-axis at A and the y-axis at B. Calculate the length AB. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (7 − 1)/(5 − 2)	✓ 1
A1 2	✓ 1
Part (b)
M1 gradient of l₁ = 3/k	✓ 1
M1 (3/k) × 2 = −1	✓ 1
A1 k = −6	✓ 1
Part (c)
M1 3 × 2 + 6 × 1 = 12, so (2, 1) is on l₁	✓ 1
R1 (2, 1) is given as a point on l₂, so it is on both lines	✓ 1
Part (d)
B1 A = (4, 0) and B = (0, 2)	✓ 1
M1 √(4² + 2²)	✓ 1
A1 4.47 km	✓ 1

Common error: Forgetting that −k with k = −6 changes the sign of the y-term: 3x + 6y = 12.

+ Add to test✓ In your test159IGMC3-3-09r12-12Gradient, midpoint and length of a line segment Paper EHard11 marks

No calculator

The corners of a glass atrium in an architect's plan are A(0, 1), B(6, 3) and C(4, 9). Each unit represents 1 m. The atrium is completed as a rectangle ABCD.

Diagram not to scale

(a)

Show that angle ABC is 90°.

(3)

(b)

Find the coordinates of D.

(2)

(c)

Find the equation of the diagonal BD in the form y = mx + c.

(3)

(d)

Find the exact area of the rectangle.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient of AB = \(\frac{2}{6}=\frac{1}{3}\)	✓ 1
M1 gradient of BC = \(\frac{6}{-2}=-3\)	✓ 1
R1 product of gradients is −1, so AB is perpendicular to BC	✓ 1
Part (b)
M1 e.g. D = A + (C − B), or midpoint of BD equals midpoint of AC	✓ 1
A1 (−2, 7)	✓ 1
Part (c)
M1 gradient \(\frac{7-3}{-2-6}=-\frac{1}{2}\)	✓ 1
M1 \(y-3=-\frac{1}{2}(x-6)\)	✓ 1
A1 \(y=-\frac{1}{2}x+6\)	✓ 1
Part (d)
M1 AB = \(\sqrt{40}\)	✓ 1
M1 BC = \(\sqrt{40}\)	✓ 1
A1 40 m²	✓ 1

Common error: Two lines are perpendicular when the product of their gradients is −1; showing the gradients are merely different is not enough.

+ Add to test✓ In your test160IGMC3-3-09r12-13Gradient, midpoint and length of a line segment Paper EHard8 marks

Calculator

Two warehouses are at A(−3, 4) and B(9, −2). Each unit on the grid represents 1 km. A new distribution centre P is to be the same distance from A and B, and it must lie on a road with equation 3x + y = 15.

Diagram not to scale

(a)

Find the equation of the perpendicular bisector of AB, in the form y = mx + c.

(3)

(b)

Find the coordinates of P.

(3)

(c)

Calculate the distance PA, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint (3, 1)	✓ 1
M1 gradient of AB is \(-\frac{1}{2}\), so perpendicular gradient is 2	✓ 1
A1 y = 2x − 5	✓ 1
Part (b)
M1 substitute y = 2x − 5 into 3x + y = 15	✓ 1
A1 x = 4	✓ 1
A1 y = 3, so P is (4, 3)	✓ 1
Part (c)
M1 \(\sqrt{(4-(-3))^2+(3-4)^2}\)	✓ 1
A1 7.07 km	✓ 1

Exam tip: Check your point P by finding PB as well; it should equal PA.

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Calculator

A survey line in a marine reserve lies along L₁: \(y=\frac{2}{3}x+1\). A dive site is at C(8, −1). Each unit on the grid represents 1 km.

Diagram not to scale

(a)

Write down the gradient of the line through C that is perpendicular to L₁.

(1)

(b)

Find the equation of this perpendicular line, in the form y = mx + c.

(2)

(c)

The perpendicular meets L₁ at F. Find the coordinates of F, correct to 3 significant figures.

(3)

(d)

Calculate the shortest distance from C to L₁, correct to 3 significant figures. Use the unrounded coordinates of F.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(-\frac{3}{2}\)	✓ 1
Part (b)
M1 \(y+1=-\frac{3}{2}(x-8)\)	✓ 1
A1 \(y=-\frac{3}{2}x+11\)	✓ 1
Part (c)
M1 equate \(\frac{2}{3}x+1=-\frac{3}{2}x+11\)	✓ 1
A1 x = 4.62 (exactly \(\frac{60}{13}\))	✓ 1
A1 y = 4.08 (exactly \(\frac{53}{13}\))	✓ 1
Part (d)
M1 distance CF = \(\sqrt{\left(8-\frac{60}{13}\right)^2+\left(-1-\frac{53}{13}\right)^2}\)	✓ 1
A1 6.10 km	✓ 1

Exam tip: The shortest distance from a point to a line is measured along the perpendicular.

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Calculator

Two water mains are modelled as straight lines on a map with units of kilometres. Main M₁ has equation 3y + 2x = 12. Main M₂ passes through the point (0, −5) and is perpendicular to M₁.

(a)

Find the gradient of M₁.

(2)

(b)

Find the equation of M₂ in the form y = mx + c.

(3)

(c)

The mains meet at the point X. Find the coordinates of X, giving each coordinate to 3 significant figures.

(3)

(d)

A pumping station is at (0, −5). Calculate its distance from X, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearrange to 3y = −2x + 12	✓ 1
A1 −2/3	✓ 1
Part (b)
B1 gradient of M₂ = 3/2 (ECF from (a))	✓ 1
M1 c = −5 from the point (0, −5)	✓ 1
A1 y = 3/2 x − 5	✓ 1
Part (c)
M1 equate −2/3 x + 4 = 3/2 x − 5	✓ 1
A1 x = 54/13 = 4.15	✓ 1
A1 y = 16/13 = 1.23	✓ 1
Part (d)
M1 distance formula with X(54/13, 16/13) (or 4.15, 1.23)	✓ 1
A1 7.49 km	✓ 1

Exam tip: If the next part needs coordinates found earlier, keep them as exact fractions or full calculator values to avoid rounding drift.

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Calculator

In a game, three towers are at A(−2, 1), B(4, 5) and C(6, 2).

Diagram not to scale

(a)

Show that AB is perpendicular to BC.

(3)

(b)

Find the exact lengths of AB and BC.

(2)

(c)

M is the midpoint of AC. Show that MB is equal to half of AC.

(3)

(d)

Find the area of triangle ABC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient AB = (5 − 1)/(4 − (−2)) = 2/3	✓ 1
M1 gradient BC = (2 − 5)/(6 − 4) = −3/2	✓ 1
A1 product 2/3 × (−3/2) = −1, so perpendicular	✓ 1
Part (b)
B1 AB = √(6² + 4²) = √52	✓ 1
B1 BC = √(2² + 3²) = √13	✓ 1
Part (c)
B1 M = (2, 1.5)	✓ 1
M1 MB² = (4 − 2)² + (5 − 1.5)² = 16.25	✓ 1
A1 AC² = 8² + 1² = 65, so (½AC)² = 16.25 = MB²	✓ 1
Part (d)
M1 ½ × √52 × √13 (ECF from (b))	✓ 1
A1 13 square units	✓ 1

Exam tip: If two sides of a triangle are perpendicular, they can serve as base and height, so the area is half their product.

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Calculator

In a game, a beam L₁ has equation y = kx + 4. A second beam L₂ passes through (1, −2) and (5, 6), and L₁ is perpendicular to L₂.

(a)

Find the gradient of L₂.

(2)

(b)

Find the value of k.

(1)

(c)

Find the equation of L₂.

(2)

(d)

Find the coordinates of the point where the beams meet.

(3)

(e)

The two beams and the y-axis form a triangle. Calculate the area of this triangle.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (6 − (−2))/(5 − 1)	✓ 1
A1 2	✓ 1
Part (b)
A1 k = −1/2 (ECF from (a))	✓ 1
Part (c)
M1 −2 = 2 × 1 + c, or y + 2 = 2(x − 1)	✓ 1
A1 y = 2x − 4	✓ 1
Part (d)
M1 −½x + 4 = 2x − 4	✓ 1
A1 x = 3.2	✓ 1
A1 y = 2.4	✓ 1
Part (e)
B1 the beams meet the y-axis at (0, 4) and (0, −4), so the base is 8	✓ 1
M1 ½ × 8 × 3.2 (the height is the x-coordinate of the meeting point)	✓ 1
A1 12.8 square units	✓ 1

Common error: Use the perpendicular distance from the meeting point to the y-axis as the height, which is its x-coordinate, not the distance along a beam.

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Two telecoms masts are at A(−1, 3) and B(7, 7) on a grid with units in kilometres. A new relay tower T must be the same distance from A and from B, and must lie on a straight road with equation \(y=x-1\).

Diagram not to scale

(a)

Find the midpoint of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AB, in the form \(y=mx+c\).

(3)

(c)

Find the coordinates of T.

(3)

(d)

Find the distance TA.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 (3, 5)	✓ 1
Part (b)
B1 gradient of AB is \(\frac{1}{2}\)	✓ 1
M1 \(y-5=-2(x-3)\)	✓ 1
A1 \(y=-2x+11\)	✓ 1
Part (c)
M1 \(-2x+11=x-1\)	✓ 1
A1 \(x=4\)	✓ 1
A1 \(y=3\), so T is (4, 3)	✓ 1
Part (d)
A1 5 km	✓ 1

Exam tip: Points equidistant from two others lie on their perpendicular bisector, which passes through the midpoint at right angles.

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Calculator

On a map grid with units in kilometres, a ridge passes through camps A(−2, −1) and B(6, 3). A climber at C(5, −3) wants to reach the ridge by the shortest possible route.

Diagram not to scale

(a)

Show that the equation of line AB is \(y=\frac{1}{2}x\).

(2)

(b)

Find the equation of the line through C that is perpendicular to AB.

(2)

(c)

Find the coordinates of the point F where this line meets AB.

(2)

(d)

Calculate the shortest distance from C to the ridge, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient \(\frac{3-(-1)}{6-(-2)}=\frac{1}{2}\)	✓ 1
A1 substitutes a point, e.g. \(\frac{1}{2}\times6=3\), so \(y=\frac{1}{2}x\)	✓ 1
Part (b)
M1 gradient \(-2\) and \(y+3=-2(x-5)\)	✓ 1
A1 \(y=-2x+7\)	✓ 1
Part (c)
M1 \(\frac{1}{2}x=-2x+7\)	✓ 1
A1 (2.8, 1.4)	✓ 1
Part (d)
M1 \(\sqrt{(5-2.8)^2+(-3-1.4)^2}\)	✓ 1
A1 4.92 km	✓ 1

Exam tip: The shortest distance from a point to a line is along the perpendicular, so find the foot of the perpendicular first.

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On a game map, a path goes from P(1, 2) to Q(9, 6). At Q the path turns through a right angle and carries on until it reaches the x-axis at R.

Diagram not to scale

(a)

Find the gradient of PQ.

(1)

(b)

Find the equation of the line QR.

(3)

(c)

Find the coordinates of R.

(2)

(d)

Calculate the area of triangle PQR.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1/2	✓ 1
Part (b)
M1 gradient of QR = −2	✓ 1
M1 y − 6 = −2(x − 9)	✓ 1
A1 y = −2x + 24	✓ 1
Part (c)
M1 substitute y = 0	✓ 1
A1 R = (12, 0)	✓ 1
Part (d)
M1 PQ = √(8² + 4²) = √80	✓ 1
M1 QR = √(3² + 6²) = √45	✓ 1
M1 ½ × √80 × √45 (right angle at Q)	✓ 1
A1 30 square units	✓ 1

Exam tip: Where a triangle has a right angle, the two sides meeting at it can be used as base and height; check perpendicularity by showing the gradients multiply to −1.

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The posts of a solar array are at A(1, 2), B(7, 4) and C(9, 10). A fourth post D is placed so that ABCD is a parallelogram. The coordinates are in metres.

Diagram not to scale

(a)

Find the midpoint of AC and use it to find the coordinates of D.

(3)

(b)

Show that AB is parallel to DC.

(2)

(c)

Show that ABCD is a rhombus.

(3)

(d)

Calculate the area of the rhombus.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 midpoint of AC = (5, 6)	✓ 1
M1 diagonals of a parallelogram bisect each other, so (7 + x)/2 = 5 and (4 + y)/2 = 6	✓ 1
A1 D = (3, 8)	✓ 1
Part (b)
M1 gradient AB = (4 − 2)/(7 − 1) = 1/3	✓ 1
A1 gradient DC = (10 − 8)/(9 − 3) = 1/3, so the lines are parallel	✓ 1
Part (c)
M1 AB = √(6² + 2²) = √40	✓ 1
M1 BC = √(2² + 6²) = √40	✓ 1
R1 a parallelogram with two adjacent sides equal is a rhombus	✓ 1
Part (d)
M1 AC = √128 and BD = √32	✓ 1
M1 area = ½ × AC × BD	✓ 1
A1 32 m²	✓ 1

Common error: The area of a rhombus is half the product of its diagonals, not half the product of its sides.

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A maintenance robot travels along a straight track l1 through A(−2, 3) and B(4, 6) on a plan where 1 unit represents 1 m. A second straight track l2 has equation y = kx + 7 and is perpendicular to l1.

Diagram not to scale

(a)

Find the gradient of l1.

(2)

(b)

Find the value of k.

(2)

(c)

Find the coordinates of the point where the two tracks cross.

(3)

(d)

The robot moves from A to l2 by the shortest route. Calculate the length of this route, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (6 − 3)/(4 − (−2))	✓ 1
A1 1/2	✓ 1
Part (b)
M1 k × ½ = −1	✓ 1
A1 k = −2	✓ 1
Part (c)
M1 equation of l1: y = ½x + 4	✓ 1
M1 equate: ½x + 4 = −2x + 7	✓ 1
A1 (1.2, 4.6)	✓ 1
Part (d)
B1 shortest route is along l1 to the crossing point, since l1 is perpendicular to l2	✓ 1
M1 √(3.2² + 1.6²)	✓ 1
A1 3.58 m	✓ 1

Exam tip: The shortest distance from a point to a line is measured along the perpendicular from the point.

+ Add to test✓ In your test170IGMC3-3-11r35-12Gradient, midpoint and length of a line segment Paper EHard9 marks

Calculator

A map of a region has 1 unit representing 1 km. Two airfields are at A(−2, 1) and B(6, 5). A rescue helicopter base H must be the same distance from A and B.

Diagram not to scale

(a)

Find the coordinates of the midpoint M of AB.

(2)

(b)

Find the gradient of AB.

(1)

(c)

Find the equation of the perpendicular bisector of AB, in the form y = mx + c.

(3)

(d)

The base H lies on the bisector and on the y-axis. Write down the coordinates of H.

(1)

(e)

Calculate the distance AH. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x = 2	✓ 1
B1 y = 3	✓ 1
Part (b)
B1 1/2	✓ 1
Part (c)
B1 gradient of the bisector −2 (ECF from (b))	✓ 1
M1 y − 3 = −2(x − 2) using M (ECF from (a))	✓ 1
A1 y = −2x + 7	✓ 1
Part (d)
B1 (0, 7) (ECF from (c))	✓ 1
Part (e)
M1 √((0 − (−2))² + (7 − 1)²) = √40	✓ 1
A1 6.32 km	✓ 1

Exam tip: Every point on the perpendicular bisector of AB is equally far from A and from B, so a quick check of AH and BH confirms your line.

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No calculator

Three navigation beacons are at A(1, 2), B(5, 4) and C(3, 8) on a grid in which 1 unit represents 1 km.

Diagram not to scale

(a)

Show that AB is perpendicular to BC.

(3)

(b)

Show that AB and BC have equal lengths.

(2)

(c)

Find the area of triangle ABC.

(2)

(d)

ABCD is a square. Find the coordinates of D.

(2)

(e)

Find the area of the square ABCD.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient of AB = 2/4 = 1/2	✓ 1
B1 gradient of BC = (8 − 4) ÷ (3 − 5) = −2	✓ 1
B1 product of gradients is −1, so AB is perpendicular to BC	✓ 1
Part (b)
M1 AB² = 4² + 2² and BC² = (−2)² + 4²	✓ 1
A1 both equal 20, so AB = BC = √20	✓ 1
Part (c)
M1 1/2 × √20 × √20	✓ 1
A1 10 km²	✓ 1
Part (d)
M1 uses the midpoint of AC, (2, 5), or the vector from B to A added to C	✓ 1
A1 D = (−1, 6)	✓ 1
Part (e)
B1 20 km² (ECF from (c))	✓ 1

Common error: Perpendicular lines have gradients whose product is −1; showing that two gradients are merely different does not prove a right angle.

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An irrigation channel on a farm plan runs in a straight line from A(0, −3) to B(6, 9). A second channel l is perpendicular to AB and passes through the midpoint M of AB. Channel l crosses the x-axis at C and the y-axis at D. O is the origin.

Diagram not to scale

(a)

Find the gradient of AB.

(2)

(b)

Find the coordinates of M.

(2)

(c)

Find the equation of l in the form y = mx + c.

(3)

(d)

Find the area of triangle OCD.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (9 − (−3)) ÷ (6 − 0)	✓ 1
A1 2	✓ 1
Part (b)
B1 x = 3	✓ 1
B1 y = 3	✓ 1
Part (c)
B1 gradient of l is −1/2 (ECF from (a))	✓ 1
M1 y − 3 = −1/2(x − 3), using M (ECF from (b))	✓ 1
A1 y = −1/2 x + 9/2	✓ 1
Part (d)
B1 C = (9, 0)	✓ 1
B1 D = (0, 4.5)	✓ 1
M1 1/2 × OC × OD (ECF from (c))	✓ 1
A1 20.25 square units	✓ 1

Exam tip: Where a line meets the axes, the triangle with the origin has the intercepts as its base and height, so no further coordinate work is needed.

+ Add to test✓ In your test173IGMC3-3-10r43-07Equation of a straight line, parallel and perpendicular lines Paper EHard9 marks

No calculator

A guide rail for a robot lies along the line L with equation y = −3x + 7. A second robot is at P(6, 5). It moves along the perpendicular to L through P until it meets L at Q.

(a)

Find the equation of the line PQ. Give your answer in the form y = mx + c.

(3)

(b)

Find the coordinates of Q.

(3)

(c)

Find the exact distance PQ.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient of PQ = 1/3	✓ 1
M1 y − 5 = (1/3)(x − 6)	✓ 1
A1 y = (1/3)x + 3	✓ 1
Part (b)
M1 −3x + 7 = (1/3)x + 3	✓ 1
A1 x = 6/5	✓ 1
A1 y = 17/5 (Q is (6/5, 17/5))	✓ 1
Part (c)
M1 PQ² = (6 − 6/5)² + (5 − 17/5)²	✓ 1
A1 (24/5)² + (8/5)² = 640/25	✓ 1
A1 PQ = 8√10 / 5 (≈ 5.06)	✓ 1

Exam tip: To meet a line at right angles, use the negative reciprocal gradient, then solve simultaneously.

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Calculator

Two charging stations for robots are at P(−1, 3) and Q(5, 7). A guide rail is laid along the perpendicular bisector of PQ. The point R lies on this rail and on the x-axis.

Diagram not to scale

(a)

Find the coordinates of the midpoint of PQ.

(2)

(b)

Find the equation of the perpendicular bisector of PQ. Give your answer in the form y = mx + c.

(3)

(c)

Find the x-coordinate of R. Give your answer to 3 significant figures.

(2)

(d)

Calculate the distance RP. Use the exact value of the x-coordinate from (c). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 x = 2	✓ 1
B1 y = 5	✓ 1
Part (b)
B1 gradient of PQ = 4/6 = 2/3	✓ 1
M1 perpendicular gradient −3/2 and line through (2, 5)	✓ 1
A1 y = −1.5x + 8	✓ 1
Part (c)
M1 0 = −1.5x + 8	✓ 1
A1 x = 16/3 = 5.33	✓ 1
Part (d)
M1 √((16/3 − (−1))² + (0 − 3)²)	✓ 1
A1 7.01	✓ 1

Exam tip: Every point on a perpendicular bisector is the same distance from both end points, so RP = RQ is a useful check.

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A fashion designer cuts a fabric panel ABCD with A(1, 2), B(7, 5), C(9, 9) and D(3, 6). Coordinates are in centimetres.

Diagram not to scale

(a)

Show that AB is parallel to DC.

(2)

(b)

Show that AD is parallel to BC.

(2)

(c)

Show that the panel is not a rectangle.

(2)

(d)

Calculate the length of the diagonal AC. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient AB = (5 − 2)/(7 − 1) = 1/2 and gradient DC = (9 − 6)/(9 − 3) = 1/2	✓ 1
A1 equal gradients, so AB is parallel to DC	✓ 1
Part (b)
M1 gradient AD = (6 − 2)/(3 − 1) = 2 and gradient BC = (9 − 5)/(9 − 7) = 2	✓ 1
A1 equal gradients, so AD is parallel to BC (ABCD is a parallelogram)	✓ 1
Part (c)
M1 gradient of AB × gradient of AD = 1/2 × 2 = 1	✓ 1
A1 the product is not −1, so AB and AD are not perpendicular	✓ 1
Part (d)
M1 √((9 − 1)² + (9 − 2)²)	✓ 1
A1 10.6 cm (√113 = 10.63…)	✓ 1

Exam tip: To show a shape is a rectangle you need both pairs of opposite sides parallel and adjacent sides perpendicular.

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On a map of a solar farm, 1 unit represents 1 km. A straight fence follows the line 3y + x = 6. A monitoring station is at P(2, 5). A cable is to be laid from P to the fence by the shortest possible route.

Diagram not to scale

(a)

Find the gradient of the fence, and hence the gradient of the cable.

(3)

(b)

Find the equation of the line along which the cable lies.

(2)

(c)

Find the coordinates of the point where the cable meets the fence.

(3)

(d)

Calculate the length of the cable. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges to y = −(1/3)x + 2	✓ 1
A1 fence gradient −1/3	✓ 1
A1 cable gradient 3 (perpendicular to the fence)	✓ 1
Part (b)
M1 uses gradient 3 and the point (2, 5), e.g. y − 5 = 3(x − 2)	✓ 1
A1 y = 3x − 1	✓ 1
Part (c)
M1 substitutes y = 3x − 1 into 3y + x = 6 to get 10x − 3 = 6	✓ 1
A1 x = 0.9	✓ 1
A1 y = 1.7	✓ 1
Part (d)
M1 √((2 − 0.9)² + (5 − 1.7)²) or √12.1	✓ 1
A1 3.48 km	✓ 1

Exam tip: The shortest distance from a point to a line is along the perpendicular. Find the foot of the perpendicular first, then use the length formula.

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Two wind turbines stand at P(−2, 5) and Q(4, 1) on a map grid where 1 unit represents 1 km. A substation S is to be built on the x-axis so that it is the same distance from P as from Q.

Diagram not to scale

(a)

Find the midpoint of PQ and the gradient of PQ.

(3)

(b)

Find the equation of the perpendicular bisector of PQ.

(3)

(c)

Find the coordinates of S.

(2)

(d)

Calculate the distance PS. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 midpoint (1, 3)	✓ 1
M1 (1 − 5)/(4 − (−2))	✓ 1
A1 −2/3	✓ 1
Part (b)
M1 perpendicular gradient 3/2 (negative reciprocal)	✓ 1
M1 substitutes (1, 3), e.g. y − 3 = 1.5(x − 1)	✓ 1
A1 y = 1.5x + 1.5 (or 2y = 3x + 3)	✓ 1
Part (c)
M1 sets y = 0 in the bisector equation	✓ 1
A1 S(−1, 0)	✓ 1
Part (d)
M1 √((−1 + 2)² + (0 − 5)²) or √26	✓ 1
A1 5.10 km	✓ 1

Exam tip: Every point on the perpendicular bisector of PQ is equidistant from P and Q. Check your answer by calculating both PS and QS.

+ Add to test✓ In your test178IGMC3-3-10r47-10Gradient, midpoint and length of a line segment Paper EHard8 marks

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Telecoms cable AB joins A(1, 2) to B(7, k). Cable CD lies along the line 2x + 3y = 12. The cables are perpendicular to each other. Units are kilometres.

(a)

Find the gradient of CD.

(2)

(b)

Find the value of k.

(3)

(c)

Find the exact length of AB.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearrange to y = −(2/3)x + 4	✓ 1
A1 −2/3	✓ 1
Part (b)
M1 gradient of AB = (k − 2)/6	✓ 1
M1 (k − 2)/6 × (−2/3) = −1	✓ 1
A1 k = 11	✓ 1
Part (c)
M1 AB = √(6² + 9²) (ECF from (b))	✓ 1
A1 √117	✓ 1
A1 3√13 km (= 10.8167…)	✓ 1

Common error: Perpendicular gradients multiply to give −1; do not just change the sign of the first gradient.

+ Add to test✓ In your test179IGMC3-3-10r47-12Gradient, midpoint and length of a line segment Paper EHard9 marks

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Four ground anchors for a radio mast are at A(−2, 1), B(4, 3), C(8, 7) and D. ABCD is a parallelogram, so the diagonals AC and BD have the same midpoint. Units are metres.

Diagram not to scale

(a)

Find the midpoint of AC.

(2)

(b)

Find the coordinates of D.

(3)

(c)

Show that AB is parallel to DC.

(2)

(d)

Calculate the length of AD. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ((−2 + 8)/2, (1 + 7)/2)	✓ 1
A1 (3, 4)	✓ 1
Part (b)
M1 midpoint of BD is the same as that of AC, so (4 + x)/2 = 3 and (3 + y)/2 = 4	✓ 1
A1 x = 2	✓ 1
A1 y = 5, so D = (2, 5) (ECF from (a))	✓ 1
Part (c)
M1 gradient AB = 2/6 = 1/3	✓ 1
A1 gradient DC = (7 − 5)/(8 − 2) = 1/3, so the lines are parallel	✓ 1
Part (d)
M1 √((2 + 2)² + (5 − 1)²) = √32 (ECF from (b))	✓ 1
A1 5.66 m	✓ 1

Common error: In a parallelogram it is the diagonals that share a midpoint, not the adjacent vertices; label the vertices in order round the shape.