ExaminerPrepexaminerprep.com · P1.4 Straight-line graphs Questions
Contact
Free access until 31 December 2026 · Physics and Maths questions and notes, updated continuously · Build a test · Get an email before anything changes

ExaminerPrep › Question bank › IAL Maths › P1.4 Straight-line graphs

Edexcel International A Level Maths · P1 · Pure Mathematics 1

Straight-line graphs

28 free, high-quality past-paper-style questions on Straight-line graphs for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 235 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1IALM-P1-4-B-01 P1 · ASEasy5 marks

The points P and Q have coordinates (−3, 7) and (9, −2) respectively.

OxyP(−3, 7)Q(9, −2)R
Figure: the line through P and Q
(a)

Find the gradient of PQ.

(2)
(b)

Find an equation of the line PQ, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(2)
(c)

The line PQ crosses the x-axis at the point R. Find the coordinates of R.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient = (−2 − 7) ÷ (9 − (−3))M1
= −9 ÷ 12 = −3⁄4A1
Part (b)
y − 7 = −3⁄4(x + 3) or y + 2 = −3⁄4(x − 9), ft their gradientM1
3x + 4y − 19 = 0 (or any non-zero integer multiple)A1
Part (c)
y = 0 gives 3x = 19, so R is (19⁄3, 0), ft their equationB1

Common error: Writing 9 − 3 instead of 9 − (−3) in the denominator. Put brackets round every negative coordinate before you subtract.

Theory: P1.4 Straight-line graphs
2IALM-P1-4-B-02 P1 · ASEasy6 marks

The line L has equation 5x − 2y + 8 = 0.

(a)

Find the gradient of L and the coordinates of the point where L crosses the y-axis.

(2)
(b)

Show that the point (−4, −6) lies on L.

(1)
(c)

Find an equation of the line through (3, 1) that is parallel to L, giving your answer in the form y = mx + c.

(2)
(d)

Write down an equation of the line through the origin that is perpendicular to L.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
2y = 5x + 8, so y = 5⁄2x + 4M1
gradient 5⁄2; crosses the y-axis at (0, 4)A1
Part (b)
5(−4) − 2(−6) + 8 = −20 + 12 + 8 = 0, so (−4, −6) lies on LB1
Part (c)
y − 1 = 5⁄2(x − 3), ft their gradientM1
y = 5⁄2x − 13⁄2A1
Part (d)
y = −2⁄5x (or 2x + 5y = 0)B1

Exam tip: Make y the subject before reading off a gradient: the coefficient of x in 5x − 2y + 8 = 0 is not the gradient.

Theory: P1.4 Straight-line graphs
3IALM-P1-4-B-03 P1 · ASEasy5 marks

The points A and B have coordinates (−5, 2) and (7, −3) respectively.

OxyA(−5, 2)B(7, −3)E
Figure: the points A, B and E (B is the midpoint of AE)
(a)

Find the coordinates of the midpoint of AB.

(1)
(b)

Find the length of AB.

(2)
(c)

The point B is the midpoint of the line segment AE. Find the coordinates of E.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
(−5 + 7⁄2, 2 + (−3)⁄2) = (1, −1⁄2)B1
Part (b)
AB = √((7 − (−5))² + (−3 − 2)²) = √(12² + 5²)M1
= √169 = 13A1
Part (c)
with E(p, q): (−5 + p⁄2, 2 + q⁄2) = (7, −3), or add the step from A to B on to BM1
E is (19, −8)A1

Common error: Finding E as the midpoint of AB again. If B is the midpoint of AE, then E is as far beyond B as A is before it: E = 2B − A.

Theory: P1.4 Straight-line graphs
4IALM-P1-4-B-04 P1 · ASEasy6 marks

The line l1 passes through the point (2, −1) and has gradient 3. The line l2 has equation x + 3y = 9.

Oxyl₁l₂(2, −1)
Figure: the lines l₁ and l₂
(a)

Find an equation of l1 in the form y = mx + c.

(1)
(b)

Show that l1 and l2 are perpendicular.

(2)
(c)

Find the coordinates of the point where l1 and l2 meet.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
y + 1 = 3(x − 2), so y = 3x − 7B1
Part (b)
y = −1⁄3x + 3, so the gradient of l2 is −1⁄3B1
3 × (−1⁄3) = −1, so the lines are perpendicularA1
Part (c)
x + 3(3x − 7) = 9M1
10x = 30, x = 3A1
y = 2: the lines meet at (3, 2)A1

Exam tip: In a “show that” about perpendicular lines, write both gradients, multiply them, get −1 and then say “so the lines are perpendicular”.

Theory: P1.4 Straight-line graphs
5IALM-P1-4-FB-01 P1 · ASEasy5 marks

Figure 1 shows the straight line ℓ. The marked points A and B lie on ℓ at gridline intersections.

xxyy−5−5−4−4−3−3−2−2−1−11122334455667788−4−4−3−3−2−2−1−1112233445566OOAABBℓℓ
(a)

Use Figure 1 to write down the coordinates of A and of B.

(1)
(b)

Find the gradient of ℓ.

(1)
(c)

Find an equation of ℓ, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(2)
(d)

The point (k, −7) lies on ℓ. Find the value of k.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
A(−3, 5) and B(6, −1), both neededB1
Part (b)
(−1 − 5) ÷ (6 − (−3)) = −6 ÷ 9 = −2⁄3, ft their pointsB1
Part (c)
y − 5 = −2⁄3(x + 3) or y + 1 = −2⁄3(x − 6), ft their gradientM1
2x + 3y − 9 = 0 (or any non-zero integer multiple)A1
Part (d)
2k − 21 − 9 = 0, so k = 15, ft their equationB1

Exam tip: Read coordinates only where the line crosses gridline intersections exactly; a point read between gridlines gives an inaccurate gradient.

Theory: P1.4 Straight-line graphs
6IALM-P1-4-FB-02 P1 · ASEasy7 marks

Two taxi companies, A and B, each charge for a journey according to its length. Figure 1 shows the cost, £C, of a journey of d km with each company. The marked points lie on gridline intersections. The cost for each company is modelled by the straight line shown.

d / kmd / kmC / £C / £224466881010121214144488121216162020242428283232363600Company BCompany BCompany ACompany A
(a)

Find the gradient of the line for Company A and explain what it represents in this context.

(2)
(b)

Write down an equation for Company A's cost in the form C = md + c.

(1)
(c)

Find an equation for Company B's cost.

(1)
(d)

Use algebra to find the length of journey for which the two companies charge the same, and that cost.

(2)
(e)

A customer plans a journey of 15 km. State, with a reason, which company is cheaper.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
(16 − 4) ÷ (8 − 0) = 1.5B1
the cost per kilometre: each extra km costs £1.50B1
Part (b)
C = 1.5d + 4B1
Part (c)
line through (0, 0) and (8, 20): C = 2.5dB1
Part (d)
1.5d + 4 = 2.5dM1
d = 4 km, cost £10A1
Part (e)
A: 1.5 × 15 + 4 = £26.50 and B: 2.5 × 15 = £37.50, so Company AB1

Exam tip: In a context the gradient is a rate (here pounds per kilometre) and the intercept is the value when the other variable is zero (here the fixed charge).

Theory: P1.4 Straight-line graphs
7IALM-P1-4-FB-03 P1 · ASEasy6 marks

Figure 1 shows the triangle ABC, with the coordinates of its vertices. M is the midpoint of AB and N is the midpoint of BC.

xxyyOOA(−4, −1)A(−4, −1)B(2, 7)B(2, 7)C(10, 1)C(10, 1)MMNN
(a)

Find the coordinates of M and of N.

(2)
(b)

Show that MN is parallel to AC.

(2)
(c)

Show that AC = 2MN.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
M is (−1, 3)B1
N is (6, 4)B1
Part (b)
gradient MN = (4 − 3) ÷ (6 − (−1)) = 1⁄7B1
gradient AC = (1 − (−1)) ÷ (10 − (−4)) = 2⁄14 = 1⁄7; the gradients are equal, so MN is parallel to ACA1
Part (c)
MN = √(7² + 1²) = √50 = 5√2M1
AC = √(14² + 2²) = √200 = 10√2 = 2 × 5√2, so AC = 2MNA1

Common error: Comparing the squared lengths 200 and 50 and concluding that AC = 4MN. If AC² = 4MN², then AC = 2MN.

Theory: P1.4 Straight-line graphs
8IALM-P1-4-FB-04 P1 · ASEasy6 marks

Figure 1 shows the parallelogram OABC, where O is the origin and A and C have the coordinates shown. The line through C and B meets the y-axis at D.

xxyyOOA(6, 2)A(6, 2)C(2, 5)C(2, 5)BBDD
(a)

Find the coordinates of B.

(1)
(b)

Find the gradient of OA.

(1)
(c)

Find an equation of the line CB, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(2)
(d)

Find the coordinates of D.

(1)
(e)

Show that the diagonals OB and AC bisect each other.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
B = (6 + 2, 2 + 5) = (8, 7)B1
Part (b)
2 ÷ 6 = 1⁄3B1
Part (c)
CB is parallel to OA, so its gradient is 1⁄3: y − 5 = 1⁄3(x − 2)M1
x − 3y + 13 = 0A1
Part (d)
x = 0 gives 3y = 13, so D is (0, 13⁄3), ft their equationB1
Part (e)
midpoint of OB = (4, 7⁄2) and midpoint of AC = (6 + 2⁄2, 2 + 5⁄2) = (4, 7⁄2): the same point, so the diagonals bisect each otherB1

Exam tip: In a parallelogram OABC the step from O to A is the same as the step from C to B, so when O is the origin, B = A + C.

Theory: P1.4 Straight-line graphs
9IALM-P1-4-B-05 P1 · ASMedium9 marks

The line segment AB joins A(−4, −1) to B(6, 3). The point C lies on the y-axis and is the same distance from A as from B.

OxyA(−4, −1)B(6, 3)MC
Figure: A, B, the midpoint M and the point C on the y-axis
(a)

Find the coordinates of the midpoint M of AB.

(1)
(b)

Find an equation of the perpendicular bisector of AB, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(3)
(c)

Explain why C lies on the perpendicular bisector of AB, and hence find the coordinates of C.

(2)
(d)

Verify that CA = CB, and find the exact length of CA.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
M is (1, 1)B1
Part (b)
gradient of AB = 4 ÷ 10 = 2⁄5B1
perpendicular gradient −5⁄2: y − 1 = −5⁄2(x − 1)M1
5x + 2y − 7 = 0A1
Part (c)
C is equidistant from A and B, and every point equidistant from A and B lies on the perpendicular bisector of ABB1
x = 0 gives 2y = 7, so C is (0, 7⁄2), ft their equationB1
Part (d)
CA² = (−4 − 0)² + (−1 − 7⁄2)² = 16 + 81⁄4 = 145⁄4M1
CB² = 6² + (3 − 7⁄2)² = 36 + 1⁄4 = 145⁄4, so CA = CBA1
CA = √145⁄2A1

Exam tip: Every point on a perpendicular bisector is the same distance from both ends of the segment — a quick check on your equation.

Theory: P1.4 Straight-line graphs
10IALM-P1-4-B-06 P1 · ASMedium8 marks

The line l has equation x + 2y − 10 = 0. It meets the x-axis at A and the y-axis at B. O is the origin.

OxylABC
Figure: the line l, the points A, B and C
(a)

Find the coordinates of A and of B.

(2)
(b)

Find the area of triangle OAB.

(1)
(c)

The line through B perpendicular to l meets the x-axis at C. Find an equation of this line in the form y = mx + c.

(2)
(d)

Find the coordinates of C.

(1)
(e)

Find the area of triangle ABC.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
A is (10, 0)B1
B is (0, 5)B1
Part (b)
½ × 10 × 5 = 25, ftB1
Part (c)
gradient of l is −½, so the perpendicular gradient is 2M1
y = 2x + 5A1
Part (d)
0 = 2x + 5: C is (−5⁄2, 0), ftB1
Part (e)
base AC = 10 − (−5⁄2) = 25⁄2 and height OB = 5M1
area = ½ × 25⁄2 × 5 = 125⁄4A1

Exam tip: When two vertices lie on the x-axis, use that side as the base: the height is then just the y-coordinate of the third vertex.

Theory: P1.4 Straight-line graphs
11IALM-P1-4-B-07 P1 · ASMedium7 marks

The points A, B and C have coordinates (1, −2), (4, 7) and (k, 19) respectively, where k is a constant.

(a)

Find an equation of the line through A and B in the form y = mx + c.

(2)
(b)

Given that A, B and C are collinear, find the value of k.

(2)
(c)

Show that AB : BC = 3 : 4.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient = (7 − (−2)) ÷ (4 − 1) = 3B1
y = 3x − 5A1
Part (b)
19 = 3k − 5M1
k = 8A1
Part (c)
AB = √(3² + 9²) = √90 = 3√10M1
BC = √(4² + 12²) = √160 = 4√10A1
AB : BC = 3√10 : 4√10 = 3 : 4A1

Exam tip: For collinear points, the line through any two of them must also pass through the third — substitute the third point into the equation.

Theory: P1.4 Straight-line graphs
12IALM-P1-4-B-08 P1 · ASMedium9 marks

On a map, distances are measured in kilometres from a fixed origin, with the x-axis pointing east and the y-axis pointing north. A straight road passes through the villages P(2, 9) and Q(10, 3). A farmhouse is at F(9, 10). A straight track is to be built from F to meet the road at right angles at the point T. Model the road and the track as straight lines and the villages and the farmhouse as points.

x (km, east)y (km, north)roadP(2, 9)Q(10, 3)F(9, 10)T
Figure: the road PQ and the track from F (not to scale)
(a)

Show that the road has equation 3x + 4y − 42 = 0.

(3)
(b)

Find an equation of the line of the track, in the form ax + by + c = 0, where a, b and c are integers.

(2)
(c)

Find the coordinates of T.

(3)
(d)

Find the length of the track.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient PQ = (3 − 9) ÷ (10 − 2) = −3⁄4B1
y − 9 = −3⁄4(x − 2)M1
4y − 36 = −3x + 6, so 3x + 4y − 42 = 0A1
Part (b)
gradient 4⁄3: y − 10 = 4⁄3(x − 9)M1
4x − 3y − 6 = 0A1
Part (c)
3x + 4y = 42 and 4x − 3y = 6M1
9x + 12y = 126 and 16x − 12y = 24, so 25x = 150M1
x = 6, y = 6: T is (6, 6)A1
Part (d)
FT = √(3² + 4²) = 5 kmB1

Exam tip: The shortest distance from a point to a straight line is along the perpendicular, so find the foot of the perpendicular first and then use the distance formula.

Theory: P1.4 Straight-line graphs
13IALM-P1-4-B-09 P1 · ASMedium8 marks

The points A, B, C and D have coordinates (−2, 3), (4, 5), (5, 2) and (−1, 0) respectively.

OxyA(−2, 3)B(4, 5)C(5, 2)D(−1, 0)
Figure: the quadrilateral ABCD
(a)

Show that AB is parallel to DC.

(2)
(b)

Show that AB is perpendicular to BC.

(2)
(c)

Explain why ABCD is a rectangle but not a square.

(2)
(d)

Find the area of ABCD.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient AB = 2 ÷ 6 = 1⁄3B1
gradient DC = (2 − 0) ÷ (5 − (−1)) = 1⁄3, equal so AB is parallel to DCB1
Part (b)
gradient BC = (2 − 5) ÷ (5 − 4) = −3B1
1⁄3 × (−3) = −1, so AB is perpendicular to BCA1
Part (c)
gradient AD = −3 = gradient BC, so both pairs of opposite sides are parallel and the angles are right angles: a rectangleB1
AB = √40 but BC = √10, so adjacent sides are not equal: not a squareB1
Part (d)
area = AB × BC = √40 × √10M1
= √400 = 20A1

Common error: Stopping after showing one pair of sides is parallel. A rectangle needs both pairs of opposite sides parallel and a right angle.

Theory: P1.4 Straight-line graphs
14IALM-P1-4-B-10 P1 · ASMedium7 marks

The line l1 has equation 2x − 3y + 4 = 0. The line l2 has equation kx + 4y − 1 = 0, where k is a constant.

(a)

Given that l1 and l2 are parallel, find the value of k.

(2)
(b)

Given instead that l1 and l2 are perpendicular, find the value of k.

(2)
(c)

Using k = 6, find the coordinates of the point of intersection of l1 and l2.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
gradients 2⁄3 and −k⁄4M1
−k⁄4 = 2⁄3, so k = −8⁄3A1
Part (b)
2⁄3 × (−k⁄4) = −1M1
k = 6A1
Part (c)
6x − 9y = −12 and 6x + 4y = 1, subtract to eliminate xM1
13y = 13, y = 1A1
2x − 3 + 4 = 0, x = −½: the point is (−½, 1)A1

Exam tip: Write each gradient in terms of k first, then form one equation: equal gradients for parallel, product −1 for perpendicular.

Theory: P1.4 Straight-line graphs
15IALM-P1-4-FB-05 P1 · ASMedium7 marks

Figure 1 shows the points A(−4, 2) and B(4, 6). The line through B perpendicular to AB meets the x-axis at C. The point D is such that ABCD is a rectangle.

xxyyOOA(−4, 2)A(−4, 2)B(4, 6)B(4, 6)CCDD
(a)

Find the gradient of AB.

(1)
(b)

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

(2)
(c)

Find the coordinates of C.

(1)
(d)

Find the area of triangle ABC.

(2)
(e)

Find the coordinates of D.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
4 ÷ 8 = 1⁄2B1
Part (b)
perpendicular gradient −2: y − 6 = −2(x − 4)M1
y = −2x + 14A1
Part (c)
y = 0 gives C (7, 0), ft their equationB1
Part (d)
AB = √(8² + 4²) = √80 = 4√5 and BC = √(3² + 6²) = √45 = 3√5M1
area = ½ × 4√5 × 3√5 = 30A1
Part (e)
D = A + (C − B) = (−4 + 3, 2 − 6) = (−1, −4)B1

Exam tip: When the right angle of a triangle is at B, the sides AB and BC are a base and its perpendicular height, so the area is ½ × AB × BC.

Theory: P1.4 Straight-line graphs
16IALM-P1-4-FB-06 P1 · ASMedium9 marks

Figure 1 shows the quadrilateral ABCD, with the coordinates of its vertices. The diagonals AC and BD meet at the point X.

xxyyOOA(−4, 0)A(−4, 0)B(2, 3)B(2, 3)C(8, 1)C(8, 1)D(−2, −4)D(−2, −4)XX
(a)

Show that AB is parallel to DC.

(2)
(b)

Show that AB : DC = 3 : 5.

(2)
(c)

Find an equation of AC and an equation of BD, each in the form ax + by + c = 0, where a, b and c are integers.

(2)
(d)

Find the coordinates of X.

(2)
(e)

Show that AX : XC = 3 : 5.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient AB = 3 ÷ 6 = 1⁄2B1
gradient DC = (1 − (−4)) ÷ (8 − (−2)) = 5⁄10 = 1⁄2; equal, so AB is parallel to DCB1
Part (b)
AB = √(6² + 3²) = √45 = 3√5 and DC = √(10² + 5²) = √125 = 5√5M1
AB : DC = 3√5 : 5√5 = 3 : 5A1
Part (c)
AC: gradient 1⁄12, x − 12y + 4 = 0B1
BD: gradient 7⁄4, 7x − 4y − 2 = 0B1
Part (d)
x = 12y − 4, so 7(12y − 4) − 4y − 2 = 0M1
80y = 30: X is (1⁄2, 3⁄8)A1
Part (e)
along AC the change in x from A to X is 1⁄2 − (−4) = 9⁄2 and from X to C is 8 − 1⁄2 = 15⁄2; 9⁄2 : 15⁄2 = 3 : 5B1

Exam tip: Along one straight line, lengths are in the same ratio as the changes in x, so you can compare lengths on a line without square roots.

Theory: P1.4 Straight-line graphs
17IALM-P1-4-FB-07 P1 · ASMedium8 marks

Figure 1 shows the square ABCD, where A and B have coordinates (1, 2) and (7, 4). The vertices C and D lie above the line AB, as shown.

xxyyOOA(1, 2)A(1, 2)B(7, 4)B(7, 4)CCDD
(a)

Find the gradient of AB.

(1)
(b)

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

(2)
(c)

Given that BC = AB, find the coordinates of C.

(3)
(d)

Find the coordinates of D.

(1)
(e)

Find the area of the square.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
2 ÷ 6 = 1⁄3B1
Part (b)
BC ⟂ AB, gradient −3: y − 4 = −3(x − 7)M1
y = −3x + 25A1
Part (c)
AB² = 6² + 2² = 40B1
C = (7 − t, 4 + 3t) on BC with t² + (3t)² = 40, or (x − 7)² + (−3x + 21)² = 40M1
t = 2 (C above AB), so C is (5, 10)A1
Part (d)
D = A + (C − B) = (−1, 8)B1
Part (e)
AB² = 40B1

Exam tip: Turning a step through 90° swaps its components and changes one sign: the step (6, 2) from A to B becomes (−2, 6) from B to C.

Theory: P1.4 Straight-line graphs
18IALM-P1-4-FB-08 P1 · ASMedium9 marks

Figure 1 shows the triangle PQR, with the coordinates of its vertices. L, M and N are the midpoints of QR, PR and PQ respectively. The lines PL, QM and RN are the medians of the triangle.

xxyyOOP(−3, −2)P(−3, −2)Q(3, 8)Q(3, 8)R(9, 0)R(9, 0)LLMMNNGG
(a)

Find the coordinates of L and of M.

(2)
(b)

Find an equation of the median PL in the form ax + by + c = 0, where a, b and c are integers.

(2)
(c)

Explain why the median QM has equation x = 3. The medians PL and QM meet at G. Find the coordinates of G.

(2)
(d)

Show that G also lies on the median RN.

(2)
(e)

Show that PG = 2GL.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
L is (6, 4)B1
M is (3, −1)B1
Part (b)
gradient = (4 − (−2)) ÷ (6 − (−3)) = 2⁄3M1
2x − 3y = 0A1
Part (c)
Q and M both have x-coordinate 3, so QM is a vertical line, x = 3B1
x = 3 in PL: 6 − 3y = 0, so G is (3, 2)B1
Part (d)
N is (0, 3); gradient RN = (3 − 0) ÷ (0 − 9) = −1⁄3, so RN is x + 3y − 9 = 0M1
3 + 3 × 2 − 9 = 0, so G lies on RNA1
Part (e)
PG = √(6² + 4²) = √52 = 2√13 and GL = √(3² + 2²) = √13, so PG = 2GLB1

Common error: Giving the gradient of QM as 0, or its equation as y = 3. A line through two points with the same x-coordinate is vertical: its equation is x = constant.

Theory: P1.4 Straight-line graphs
19IALM-P1-4-FB-09 P1 · ASMedium9 marks

Figure 1 shows part of a map of a hillside, with the x-axis pointing east, the y-axis pointing north and 1 unit representing 100 m. A cable car runs in a straight line from the lower station A to the upper station B. A straight road passes through the points E and F. The points A, B, E and F lie at gridline intersections. Model the cable and the road as straight lines on the map and the stations as points.

x (east)x (east)y (north)y (north)224466881010121222446688101000AABBEEFFroadroadcablecable1 unit = 100 m1 unit = 100 m
(a)

Use Figure 1 to write down the coordinates of A and of B.

(1)
(b)

Find an equation of the line AB in the form ax + by + c = 0, where a, b and c are integers.

(2)
(c)

Two pylons support the cable. On the map they divide AB into three equal parts. Find the coordinates of the two pylons.

(2)
(d)

On the map, the cable crosses the road at the point X. Find the coordinates of X, and find the distance AX on the map in metres, to 3 significant figures.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
A(1, 2) and B(13, 8), both neededB1
Part (b)
gradient = (8 − 2) ÷ (13 − 1) = 1⁄2: y − 2 = 1⁄2(x − 1)M1
x − 2y + 3 = 0A1
Part (c)
one third of the step from A to B is (12 ÷ 3, 6 ÷ 3) = (4, 2)M1
pylons at (5, 4) and (9, 6)A1
Part (d)
road through E(2, 11) and F(12, 1): gradient −1, x + y − 13 = 0B1
x = 2y − 3 substituted: 2y − 3 + y = 13M1
y = 16⁄3: X is (23⁄3, 16⁄3)A1
AX = √((20⁄3)² + (10⁄3)²) = 7.454 units, so 745 mA1

Exam tip: Work in map units all the way through and convert to metres only at the end, so that no rounding creeps in early.

Theory: P1.4 Straight-line graphs
20IALM-P1-4-FB-10 P1 · ASMedium8 marks

Figure 1 shows the lines l1: y = x + 2, l2: 2x + y − 11 = 0 and l3: x − 4y − 1 = 0. The lines enclose the shaded triangle ABC, where A is the point of intersection of l1 and l2, B is the point of intersection of l1 and l3, and C is the point of intersection of l2 and l3.

xxyy−5−5−4−4−3−3−2−2−1−111223344556677−3−3−2−2−1−111223344556677OOl₁l₁l₂l₂l₃l₃AABBCC
(a)

Use algebra to find the coordinates of A.

(2)
(b)

Use algebra to find the coordinates of B and of C.

(3)
(c)

Find the area of triangle ABC.

(2)
(d)

Show that triangle ABC has no right angle.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
2x + (x + 2) − 11 = 0M1
x = 3: A is (3, 5)A1
Part (b)
B: x − 4(x + 2) − 1 = 0 gives x = −3, so B is (−3, −1)B1
C: x = 4y + 1 in l2: 2(4y + 1) + y − 11 = 0M1
y = 1: C is (5, 1)A1
Part (c)
enclosing rectangle 8 × 6 = 48, minus the corner triangles ½ × 6 × 6 + ½ × 8 × 2 + ½ × 2 × 4 = 30 (or any valid method)M1
area = 18A1
Part (d)
the gradients are 1, −2 and 1⁄4; the products of pairs are −2, 1⁄4 and −1⁄2, none equal to −1B1

Exam tip: For a triangle whose sides all slope, draw the rectangle round it with sides parallel to the axes and subtract the right-angled triangles in the corners.

Theory: P1.4 Straight-line graphs
21IALM-P1-4-B-11 P1 · ASHard11 marks

The triangle ABC has vertices A(−3, −1), B(7, 4) and C(0, 8).

OxyA(−3, −1)B(7, 4)C(0, 8)N
Figure: triangle ABC with the perpendicular from C to AB
(a)

Find an equation of the line AB in the form x + py + q = 0, where p and q are integers.

(3)
(b)

The line l passes through C and is perpendicular to AB. Find an equation of l in the form y = mx + c.

(2)
(c)

The line l meets AB at the point N. Find the coordinates of N.

(3)
(d)

Hence find the exact area of triangle ABC.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient AB = 5 ÷ 10 = ½B1
y + 1 = ½(x + 3)M1
x − 2y + 1 = 0A1
Part (b)
gradient −2M1
y = −2x + 8A1
Part (c)
x − 2(−2x + 8) + 1 = 0M1
5x − 15 = 0, x = 3A1
N is (3, 2)A1
Part (d)
CN = √(3² + 6²) = √45 = 3√5M1
AB = √(10² + 5²) = √125 = 5√5A1
area = ½ × 5√5 × 3√5 = 75⁄2A1

Exam tip: CN is perpendicular to AB, so it is the height of the triangle when AB is the base. That is why the question asks for N first.

Theory: P1.4 Straight-line graphs
22IALM-P1-4-B-12 P1 · ASHard11 marks

ABCD is a rhombus. The opposite vertices A and C have coordinates (1, 2) and (7, 6) respectively. The vertex B lies on the y-axis.

OxyA(1, 2)C(7, 6)BD
Figure: the rhombus ABCD
(a)

Find an equation of the perpendicular bisector of AC, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(4)
(b)

Explain why B lies on this line, and find the coordinates of B.

(2)
(c)

Find the coordinates of D.

(2)
(d)

Find the area of the rhombus.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
midpoint of AC is (4, 4)B1
gradient AC = 4 ÷ 6 = 2⁄3B1
gradient −3⁄2: y − 4 = −3⁄2(x − 4)M1
3x + 2y − 20 = 0A1
Part (b)
the diagonals of a rhombus bisect each other at right angles (or: AB = CB), so B is on the perpendicular bisector of ACB1
x = 0: B is (0, 10)B1
Part (c)
the midpoint of BD is (4, 4)M1
D is (8, −2)A1
Part (d)
AC = √52 = 2√13 and BD = √208 = 4√13B1
area = ½ × AC × BD = ½ × 2√13 × 4√13M1
= 52A1

Exam tip: In a rhombus the diagonals are perpendicular bisectors of each other, so the area is half the product of the diagonals.

Theory: P1.4 Straight-line graphs
23IALM-P1-4-B-13 P1 · ASHard12 marks

The points P, Q and R have coordinates (−1, 4), (6, 5) and (7, −2) respectively.

OxyP(−1, 4)Q(6, 5)R(7, −2)
Figure: triangle PQR
(a)

Show that angle PQR is a right angle.

(2)
(b)

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

(3)
(c)

The perpendicular bisector of QR has equation x − 7y + 4 = 0. The two perpendicular bisectors meet at K. Find the coordinates of K.

(3)
(d)

Show that K is the midpoint of PR, and find the length KP.

(2)
(e)

Find the area of triangle PQR.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient PQ = 1⁄7 and gradient QR = −7B1
1⁄7 × (−7) = −1, so PQ is perpendicular to QR and angle PQR = 90°A1
Part (b)
midpoint of PQ is (5⁄2, 9⁄2)B1
y − 9⁄2 = −7(x − 5⁄2)M1
y = −7x + 22A1
Part (c)
x − 7(−7x + 22) + 4 = 0M1
50x − 150 = 0, x = 3A1
K is (3, 1)A1
Part (d)
midpoint of PR = (−1 + 7⁄2, 4 − 2⁄2) = (3, 1) = KB1
KP = √(4² + 3²) = 5B1
Part (e)
PQ = √50 = 5√2 and QR = √50 = 5√2M1
area = ½ × 5√2 × 5√2 = 25A1

Exam tip: In a right-angled triangle the midpoint of the hypotenuse is the same distance from all three vertices, so KP = KQ = KR = 5 is a quick check on your coordinates of K.

Theory: P1.4 Straight-line graphs
24IALM-P1-4-B-14 P1 · ASHard14 marks

The line l1 passes through the points A(−2, 4) and B(6, 0). The line l2 is parallel to l1 and passes through the point C(4, 5). O is the origin.

Oxyl₁l₂A(−2, 4)B(6, 0)C(4, 5)DEFGH
Figure: the parallel lines l₁ and l₂ and the quadrilateral BDEF
(a)

Find an equation of l1, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(3)
(b)

Find an equation of l2, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(2)
(c)

The line l2 meets the x-axis at D and the y-axis at E. The line l1 meets the y-axis at F. Find the area of the quadrilateral BDEF.

(4)
(d)

The line through O perpendicular to l1 meets l1 at G and l2 at H. Find the coordinates of G and of H.

(3)
(e)

Hence find the exact distance between the lines l1 and l2.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient = (0 − 4) ÷ (6 − (−2)) = −½B1
y − 0 = −½(x − 6)M1
x + 2y − 6 = 0A1
Part (b)
x + 2y + d = 0 with 4 + 10 + d = 0 (or y − 5 = −½(x − 4))M1
x + 2y − 14 = 0A1
Part (c)
D(14, 0) and E(0, 7)B1
F(0, 3)B1
area = triangle ODE − triangle OBF = ½ × 14 × 7 − ½ × 6 × 3M1
= 49 − 9 = 40A1
Part (d)
the line is y = 2xB1
x + 4x = 6 and x + 4x = 14M1
G (6⁄5, 12⁄5) and H (14⁄5, 28⁄5)A1
Part (e)
GH = √((8⁄5)² + (16⁄5)²) = √(320⁄25)M1
= 8√5⁄5A1

Common error: Measuring the distance between parallel lines along an axis (for example from F to E, which is 4). The distance between parallel lines is measured along a common perpendicular.

Theory: P1.4 Straight-line graphs
25IALM-P1-4-FB-11 P1 · ASHard10 marks

Figure 1 shows a plane mirror lying along the line ℓ with equation y = x − 1. A ray of light from the point S(1, 6) strikes the mirror at R and is reflected so that it passes through the point T(2, 10). The image S′ of S in the mirror is the point such that ℓ is the perpendicular bisector of SS′. You may assume that the reflected ray RT lies on the straight line through S′ and T.

xxyyOOS(1, 6)S(1, 6)T(2, 10)T(2, 10)RRℓℓ
(a)

Find an equation of the line through S perpendicular to ℓ, in the form y = mx + c.

(2)
(b)

Find the coordinates of the point where this line meets ℓ, and hence show that S′ is the point (7, 0).

(3)
(c)

Find an equation of the line through S′ and T, and hence find the coordinates of R.

(3)
(d)

Find the exact total length of the path SRT of the ray.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient of ℓ is 1, so the perpendicular gradient is −1M1
y = −x + 7A1
Part (b)
x − 1 = −x + 7M1
the lines meet at (4, 3)A1
(4, 3) is the midpoint of SS′, so S′ = (2 × 4 − 1, 2 × 3 − 6) = (7, 0)A1
Part (c)
gradient = (10 − 0) ÷ (2 − 7) = −2, so y = −2x + 14B1
−2x + 14 = x − 1M1
R is (5, 4)A1
Part (d)
SR = √(4² + 2²) = 2√5 and RT = √(3² + 6²) = 3√5, or SR + RT = S′T = √(5² + 10²)M1
total = 5√5A1

Exam tip: To reflect a point in a line, find the foot of the perpendicular from the point to the line. The foot is the midpoint of the point and its image, so the image is the same step again beyond the foot.

Theory: P1.4 Straight-line graphs
26IALM-P1-4-FB-12 P1 · ASHard10 marks

Figure 1 shows the triangle ABC, with the coordinates of its vertices. An altitude of a triangle is the line through a vertex perpendicular to the opposite side. The three altitudes of triangle ABC are shown dashed, and they meet at the point H.

xxyyOOA(−4, −2)A(−4, −2)B(8, 1)B(8, 1)C(2, 8)C(2, 8)HH
(a)

Find an equation of the altitude through C, in the form ax + by + c = 0, where a, b and c are integers.

(3)
(b)

Find an equation of the altitude through A, in the form ax + by + c = 0, where a, b and c are integers.

(3)
(c)

Find the coordinates of H.

(2)
(d)

Show that H lies on the altitude through B.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient AB = 3 ÷ 12 = 1⁄4B1
perpendicular gradient −4: y − 8 = −4(x − 2)M1
4x + y − 16 = 0A1
Part (b)
gradient BC = (8 − 1) ÷ (2 − 8) = −7⁄6B1
perpendicular gradient 6⁄7: y + 2 = 6⁄7(x + 4)M1
6x − 7y + 10 = 0A1
Part (c)
y = 16 − 4x: 6x − 7(16 − 4x) + 10 = 0M1
34x = 102: H is (3, 4)A1
Part (d)
gradient AC = 10 ÷ 6 = 5⁄3 and gradient BH = (4 − 1) ÷ (3 − 8) = −3⁄5M1
5⁄3 × (−3⁄5) = −1, so BH is perpendicular to AC and H lies on the altitude through BA1

Exam tip: To show that a point lies on a third line, either substitute it into that line's equation or show that the segment from the vertex to the point has the required gradient.

Theory: P1.4 Straight-line graphs
27IALM-P1-4-FB-13 P1 · ASHard11 marks

The line l has gradient −m, where m > 0, and passes through the point P(4, 3). The line l meets the x-axis at A and the y-axis at B, and O is the origin. Figure 1 shows one such line.

xxyyOOP(4, 3)P(4, 3)AABBgradient −mgradient −m
(a)

Show that A has coordinates (4 + 3⁄m, 0), and find the coordinates of B in terms of m.

(2)
(b)

Show that the area of triangle OAB is (4m + 3)2⁄2m.

(2)
(c)

Given that the area of triangle OAB is 27, find the two possible values of m.

(3)
(d)

For the larger value of m, find an equation of l in the form ax + by + c = 0, where a, b and c are integers, and the coordinates of A and B.

(2)
(e)

For each value of m, find the ratio AP : PB.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
y − 3 = −m(x − 4); y = 0 gives x − 4 = 3⁄m, so A is (4 + 3⁄m, 0)B1
x = 0 gives B (0, 3 + 4m)B1
Part (b)
½ × (4 + 3⁄m)(3 + 4m) = ½ × 4m + 3⁄m × (4m + 3)M1
= (4m + 3)2⁄2mA1
Part (c)
(4m + 3)2 = 54mM1
16m2 − 30m + 9 = 0A1
(8m − 3)(2m − 3) = 0: m = 3⁄8 or m = 3⁄2A1
Part (d)
3x + 2y − 18 = 0B1
A (6, 0) and B (0, 9)B1
Part (e)
m = 3⁄2: from A(6, 0) to B(0, 9), P is reached after a change in x of 2 out of 6, so AP : PB = 1 : 2B1
m = 3⁄8: A(12, 0), B(0, 9⁄2), change in x of 8 out of 12, so AP : PB = 2 : 1B1

Common error: Writing the x-intercept as 4 − 3/m. With a negative gradient, moving down 3 from P to the x-axis means moving right, so the x-intercept is 4 + 3/m.

Theory: P1.4 Straight-line graphs
28IALM-P1-4-FB-14 P1 · ASHard12 marks

Figure 1 shows a field OABC, where O is the origin. The field is bounded by the positive x- and y-axes and by two straight hedges. One hedge lies along l1: 2x + y − 18 = 0 and meets the x-axis at A. The other lies along l2: x − 2y + 6 = 0 and meets the y-axis at C. The hedges meet at B. One unit on each axis represents 10 m.

xxyyOOl₁l₁l₂l₂AABBCCfieldfield1 unit = 10 m1 unit = 10 m
(a)

Find the coordinates of A, B and C.

(4)
(b)

Show that the hedges meet at right angles.

(1)
(c)

Find the area of the field in m2.

(2)
(d)

A straight fence is to be built from O to a point D on the boundary of the field, dividing the field into two parts of equal area. Explain why D lies on AB, and find the coordinates of D and an equation of the line of the fence.

(4)
(e)

Find the length of the fence in metres, to 3 significant figures.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
A (9, 0)B1
C (0, 3)B1
x = 2y − 6 in l1: 2(2y − 6) + y − 18 = 0M1
B (6, 6)A1
Part (b)
gradients −2 and 1⁄2; −2 × 1⁄2 = −1, so the hedges are perpendicularB1
Part (c)
triangle OAB + triangle OBC = ½ × 9 × 6 + ½ × 3 × 6 = 27 + 9 = 36 square unitsM1
1 square unit = 100 m2, so the area is 3600 m2A1
Part (d)
triangle OBC has area 9, less than half the field (18), so D must lie on ABB1
triangle OAD: ½ × 9 × yD = 18M1
yD = 4, so D is (7, 4)A1
fence: 4x − 7y = 0 (or y = 4⁄7x)A1
Part (e)
√(7² + 4²) = √65 units, so 80.6 mB1

Exam tip: Split a quadrilateral with a vertex at the origin into two triangles from O: a triangle with one side on an axis has area ½ × that side × the other coordinate of its third vertex.

Theory: P1.4 Straight-line graphs
© ExaminerPrep · examinerprep.com · Original material by Evan Kotronis. ExaminerPrep is an independent tutoring service. It has been developed independently from and is not endorsed by the International Baccalaureate Organization. "International Baccalaureate", "IB" and "IB Diploma Programme" are registered trademarks of the IBO, used here for descriptive purposes only. Pearson Edexcel and Cambridge are trademarks of their owners; ExaminerPrep is not affiliated with any exam board.