IGCSE · Additional Mathematics 0606 · Geometry & trigonometry

Straight-line graphs: IGCSE Additional Mathematics questions with answers

Free Straight-line graphs questions for IGCSE Additional Mathematics (0606), part of Geometry & trigonometry. Each question has the answer and the marking points an examiner would look for.

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Straight-line graphs practice questions

Question 1Structured · 5 marksA* challenge

The points A(−2, 1), B(6, 5) and C(4, k) are such that angle ABC = 90°.

(a)

Find the value of k.

[3]
(b)

Find the area of triangle ABC.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)k = 9
gradient AB = ½ B1
gradient BC = −2: (k − 5) ÷ (4 − 6) = −2 M1
k = 9 A1
3
(b)20
AB = √80, BC = √20 M1
½ × √80 × √20 = 20 A1
2
Question 2Structured · 4 marks

A is (−1, 0) and B is (5, 2).

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.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
6x + 2y = 14
midpoint (2, 1) B1
gradient AB = 13 B1
perpendicular gradient −3 M1
6x + 2y = 14 (or a multiple) A1
4
Question 3Structured · 1 mark

State what is meant by the term gradient of perpendicular lines.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
Two lines are perpendicular when the product of their gradients is −1
Two lines are perpendicular when the product of their gradients is −1 B1
1
Question 4Structured · 4 marks

A is (−2, 2) and B is (0, −4).

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.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
2x − 6y = 4
midpoint (−1, −1) B1
gradient AB = −3 B1
perpendicular gradient 13 M1
2x − 6y = 4 (or a multiple) A1
4
Question 5Structured · 4 marks

A is (−3, 1) and B is (1, −1).

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.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
4x − 2y = −4
midpoint (−1, 0) B1
gradient AB = −12 B1
perpendicular gradient 2 M1
4x − 2y = −4 (or a multiple) A1
4
Question 6Structured · 4 marks

A is (−3, −1) and B is (−1, 3).

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.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
2x + 4y = 0
midpoint (−2, 1) B1
gradient AB = 2 B1
perpendicular gradient −12 M1
2x + 4y = 0 (or a multiple) A1
4

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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in