IGCSE Maths 0580 Perpendicular Lines Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Coordinate Geometry · 3.7

Every perpendicular lines question from 5 Cambridge IGCSE Mathematics (0580) past papers (Oct/Nov 2025 to May/June 2026), syllabus section 3.7. 6 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/43/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Coordinate Geometry past paper questions.

0580 Perpendicular Lines past paper questions: Core (Papers 1 and 3)

13 marksCore

Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q4

Point P is shown on the grid.

0580/32 October/November 2025 Q4 – point P on a coordinate grid from −4 to 4xy-4-3-2-11234-4-3-2-112340P

(a) Write down the coordinates of point P.

(b) On the grid, draw the line y = x.

(c) On the grid, draw the line that goes through point P and is perpendicular to the line y = x.

Show solution
Part (a)
  1. Read across (x) first, then up (y). P is 2 to the left and 3 up.
  2. (−2, 3)
Answer (−2, 3)
Part (b)
  1. On y = x, the y-coordinate always equals the x-coordinate: (−4, −4), (0, 0), (4, 4).
  2. Plot these points and rule a line through them.
Part (c)
  1. y = x has gradient 1. A perpendicular line has gradient −1 ÷ 1 = −1.
  2. The line is y = −x + c. It goes through P(−2, 3): 3 = −(−2) + c, so 3 = 2 + c and c = 1.
  3. Draw y = −x + 1, which passes through (−2, 3), (0, 1) and (1, 0).
0580/32 October/November 2025 Q4 solution – lines y = x and y = −x + 1 drawnxy-4-3-2-11234-4-3-2-112340y = xy = −x + 1P
Answer y = −x + 1

0580 Perpendicular Lines past paper questions: Extended (Papers 2 and 4)

24 marksExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q14(b)

A is the point (5, 2) and B is the point (−3, 6). Find the equation of the line perpendicular to AB that passes through the point (−2, 5). Give your answer in the form y = mx + c.

Show solution
  1. Gradient of AB = change in y⁄change in x = 6 − 2⁄−3 − 5 = 4⁄−8 = −1⁄2.
  2. A perpendicular line has the negative reciprocal gradient: −1 ÷ (−1⁄2) = 2.
  3. The line is y = 2x + c. Substitute (−2, 5): 5 = 2 × (−2) + c, so 5 = −4 + c and c = 9.
  4. y = 2x + 9
Answer y = 2x + 9
33 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q7

Line L has equation y = 2.

(a) Find the equation of the line perpendicular to line L that passes through the point (−1, 5).

(b) Find the coordinates of the point where line L intersects the line y = 20 − 6x.

Show solution
Part (a)
  1. y = 2 is a horizontal line. A line perpendicular to it is vertical, with equation x = a number.
  2. It passes through (−1, 5), so its x-coordinate is always −1.
  3. x = −1
Answer x = −1
Part (b)
  1. At the intersection, both equations are true, so y = 2 in y = 20 − 6x.
  2. 2 = 20 − 6x, so 6x = 18 and x = 3.
  3. Intersection = (3, 2)
Answer (3, 2)
49 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q19(b,c)

Point A has coordinates (−4, 3) and point B has coordinates (4, 7).

(b) Point C has coordinates (x, 7) and AC = 4√10. Find the two possible values of x.

(c) Point D lies on the line AB with AD : DB = 3 : 1. Find the equation of the line perpendicular to AB that passes through point D. Give your answer in the form y = mx + c.

Show solution
Part (b)
  1. Use the distance formula: AC² = (x − (−4))² + (7 − 3)² = (x + 4)² + 16.
  2. AC = 4√10, so AC² = 16 × 10 = 160.
  3. (x + 4)² + 16 = 160, so (x + 4)² = 144.
  4. Square root: x + 4 = 12 or x + 4 = −12.
  5. x = 8 or x = −16
Answer x = 8 or x = −16
Part (c)
  1. AD : DB = 3 : 1, so D is 3⁄4 of the way from A to B.
  2. From A to B the x changes by 8 and the y by 4. 3⁄4 of these is 6 and 3.
  3. D = (−4 + 6, 3 + 3) = (2, 6).
  4. Gradient of AB = 4⁄8 = 1⁄2. A perpendicular line has gradient −1 ÷ 1⁄2 = −2.
  5. Substitute D(2, 6) into y = −2x + c: 6 = −4 + c, so c = 10.
  6. y = −2x + 10
Answer y = −2x + 10
511 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q17

B is the point (−3, 1) and D is the point (−5, 9). BD is a diagonal of the kite ABCD.

(a) The ratio of the lengths of the diagonals BD : AC = 2 : 3. Work out the length of AC. Give your answer as a surd in its simplest form.

(b) Find the coordinates of the midpoint of BD.

(c) The diagonal AC of the kite passes through the midpoint of BD. Find an equation of AC in the form y = mx + c.

Show solution
Part (a)
  1. Length of BD using the distance formula: √((−5 − (−3))² + (9 − 1)²) = √((−2)² + 8²) = √(4 + 64) = √68.
  2. Simplify: √68 = √(4 × 17) = 2√17.
  3. BD : AC = 2 : 3, so AC = 3⁄2 × BD = 3⁄2 × 2√17 = 3√17
Answer 3√17
Part (b)
  1. Midpoint = average of the x-coordinates and of the y-coordinates.
  2. x: (−3 + (−5)) ÷ 2 = −4. y: (1 + 9) ÷ 2 = 5.
  3. Midpoint = (−4, 5)
Answer (−4, 5)
Part (c)
  1. Gradient of BD = 9 − 1⁄−5 − (−3) = 8⁄−2 = −4.
  2. The diagonals of a kite are perpendicular, so the gradient of AC is the negative reciprocal: 1⁄4.
  3. AC passes through (−4, 5): 5 = 1⁄4 × (−4) + c, so 5 = −1 + c and c = 6.
  4. y = 1⁄4x + 6
Answer y = 1⁄4x + 6
63 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q22

The equation of line L is y = −1⁄2x + 7. Find an equation of the line perpendicular to line L that passes through the point (3, 5). Give your answer in the form y = mx + c.

Show solution
  1. The gradient of L is −1⁄2. A perpendicular line has the negative reciprocal gradient: −1 ÷ (−1⁄2) = 2.
  2. The new line is y = 2x + c. It passes through (3, 5): 5 = 2 × 3 + c, so c = −1.
  3. y = 2x − 1
Answer y = 2x − 1

Question papers used on this page

All perpendicular lines questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/43/M/J/260580_s26_qp_43May/June 2026 · Paper 4 (Extended)Q14(b)
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q7, Q19(b,c)
0580/22/O/N/250580_w25_qp_22Oct/Nov 2025 · Paper 2 (Extended)Q17
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q22
0580/32/O/N/250580_w25_qp_32Oct/Nov 2025 · Paper 3 (Core)Q4

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