Original exam-style questions with full mark schemes. Use the filters to choose level and difficulty, and tick questions to build a printable test.
28questions
159marks
Fullmark schemes
Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
Make your own test. Tap + Add to test on any question, from any topic, then “Build & print test”. Or let the test builder choose questions for you in seconds.
1IGMC-043-00-01Gradient, midpoint and length of a line segmentPaper CEasy4 marks
No calculator
The point A has coordinates (2, 5) and the point B has coordinates (8, 1).
(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.
2IGMC-043-00-02Gradient, midpoint and length of a line segmentPaper CEasy5 marks
Calculator
The point P has coordinates (−1, 3) and the point Q has coordinates (6, 8).
(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.
3IGMC-043-00-03Gradient, midpoint and length of a line segmentPaper 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.
4IGMC-043-00-04Gradient, midpoint and length of a line segmentPaper 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.
5IGMC-043-00-05Gradient, midpoint and length of a line segmentPaper CEasy5 marks
Calculator
The point C has coordinates (1.5, 2.8) and the point D has coordinates (4.7, −3.6).
(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.
6IGMC-043-01-01Equation of a straight line, parallel and perpendicular linesPaper 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.
7IGMC-043-01-02Equation of a straight line, parallel and perpendicular linesPaper 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.
8IGMC-043-01-04Equation of a straight line, parallel and perpendicular linesPaper 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.
9IGMC-043-00-07Gradient, midpoint and length of a line segmentPaper CMedium9 marks
No calculator
A triangle has vertices A(1, 1), B(7, 3) and C(3, 7).
(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.
10IGMC-043-00-09Gradient, midpoint and length of a line segmentPaper CMedium4 marks
Calculator
A straight line passes through A(−2, −1) and B(5, 4).
(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.
11IGMC-043-00-10Gradient, midpoint and length of a line segmentPaper 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.
12IGMC-043-01-07Equation of a straight line, parallel and perpendicular linesPaper 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.
13IGMC-043-01-11Equation of a straight line, parallel and perpendicular linesPaper 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.
14IGMC-043-01-03Equation of a straight line, parallel and perpendicular linesPaper 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.
15IGMC-043-01-05Equation of a straight line, parallel and perpendicular linesPaper 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.
16IGMC-043-00-06Gradient, midpoint and length of a line segmentPaper 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: \(80 = 16 \times 5\).
17IGMC-043-00-08Gradient, midpoint and length of a line segmentPaper 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.
18IGMC-043-00-11Gradient, midpoint and length of a line segmentPaper 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.
19IGMC-043-01-06Equation of a straight line, parallel and perpendicular linesPaper EMedium8 marks
No calculator
The points A and B have coordinates A(−3, 4) and B(5, −2).
(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.
20IGMC-043-01-08Equation of a straight line, parallel and perpendicular linesPaper 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.
21IGMC-043-01-09Equation of a straight line, parallel and perpendicular linesPaper 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.
22IGMC-043-01-12Equation of a straight line, parallel and perpendicular linesPaper EMedium5 marks
No calculator
A quadrilateral has vertices A(1, 2), B(5, 4), C(6, 8) and D(2, 6).
(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.
23IGMC-043-00-12Gradient, midpoint and length of a line segmentPaper 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.
24IGMC-043-00-13Gradient, midpoint and length of a line segmentPaper 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.
(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.
25IGMC-043-00-14Gradient, midpoint and length of a line segmentPaper EHard8 marks
No calculator
The line segment AB joins A(2, −1) to B(10, 5).
(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.
26IGMC-043-01-10Equation of a straight line, parallel and perpendicular linesPaper EHard8 marks
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.
(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.
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.
27IGMC-043-01-13Equation of a straight line, parallel and perpendicular linesPaper EHard8 marks
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 \(\frac{2}{5}\)
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.
28IGMC-043-01-14Equation of a straight line, parallel and perpendicular linesPaper 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.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.