Free Perpendicular lines questions for IGCSE Mathematics (0580), Extended tier (Supplement content), part of Coordinate geometry. Each question has the answer and the marking points an examiner would look for.
Generate a full Perpendicular Lines worksheet →All Coordinate geometry questionsThe line L has gradient 3 and passes through the point (2, 11).
L crosses the x-axis at A and the y-axis at B. O is the origin.
Find the area of triangle OAB.
| Part | Answer and marking points | Marks |
|---|---|---|
| 256 y = 3x + 5 M1 B = (0, 5), A = (−53, 0) A1 ½ × base × height M1 256 A1 | 4 |
The line L has equation y = 2x.
A is the point (1, 7) and B is the point (3, 9). A and B are on the same side of L.
Find the coordinates of the point on L that is nearest to A, and the exact shortest distance from A to L.
P is a point on L. Find the coordinates of P for which AP + PB is as small as possible, and the exact value of this least distance.
Q is a point on L. Find the coordinates of Q for which QA − QB is as large as possible, and the exact value of this greatest difference.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | (3, 6); √5 line through A perpendicular to L: y − 7 = −½(x − 1) M1 meets y = 2x at (3, 6) A1 √(22 + 12) = √5 A1 | 3 |
| (b) | P = (3.75, 7.5); least AP + PB = 2√5 reflects A in L: A′ = (5, 5) M1 A′ = 2(3, 6) − (1, 7) A1 least AP + PB is when P is on A′B: y = −2x + 15 M1 P = (3.75, 7.5) A1 A′B = √(22 + 42) = 2√5 A1 | 5 |
| (c) | Q = (6, 12); 2√2 QA − QB ≤ AB, with equality when Q, B, A are collinear: line AB is y = x + 6 M1 Q = (6, 12) A1 AB = 2√2 A1 | 3 |
The diagram shows the line L.
Find the gradient of line L.
Find the equation of line L in the form y = mx + c.
Find the equation of the line parallel to L that passes through the point (−3, 1).
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | −23 −46 using two points on the line M1 −23 A1 | 2 |
| (b) | y = −23x y = −23x B2 B1 for correct gradient or correct intercept in y = mx + c | 2 |
| (c) | y = −23x − 1 1 = −23 × −3 + c M1 y = −23x − 1 A1 | 2 |
The equation of a line is 3x + 4y = 12.
Find the gradient of the line.
Find the coordinates of the points where the line crosses the x-axis and the y-axis.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | −34 rearranges to y = … M1 −34 A1 | 2 |
| (b) | (4, 0) and (0, 3) (4, 0) B1 (0, 3) B1 | 2 |
A is the point (−4, 1) and B is the point (−2, 5).
Find the equation of the perpendicular bisector of AB. Give your answer in the form ax + by = c, where a, b and c are integers.
The perpendicular bisector crosses the x-axis at P. Find the x-coordinate of P.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | x + 2y = 3 midpoint (−3, 3) B1 gradient AB = 2 B1 perpendicular gradient = −12 M1 substitute midpoint into y = mx + c (or equivalent) M1 x + 2y = 3 oe A1 | 5 |
| (b) | 3 3 B1 | 1 |
A is the point (4, 4) and B is the point (8, 2).
Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c.
The perpendicular bisector crosses the x-axis at P and the y-axis at Q. Find the area of triangle OPQ, where O is the origin.
Show that the point (4, −1) is equidistant from A and B.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | y = 2x − 9 midpoint (6, 3) B1 gradient AB = −12 B1 perpendicular gradient 2 M1 y = 2x − 9 A1 | 4 |
| (b) | 814 Q = (0, −9) B1 P = (92, 0) B1 814 A1 | 3 |
| (c) | it lies on the perpendicular bisector / distances equal substitutes into the equation or finds both distances M1 convincing conclusion A1 | 2 |
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Meet Riyaz →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