Every equations of linear graphs question from 5 Cambridge IGCSE Mathematics (0580) past papers (Oct/Nov 2025 to May/June 2026), syllabus section 3.5. 5 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/11/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 Equations of Linear Graphs past paper questions: Core (Papers 1 and 3)
May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q20
The diagram shows line L on a grid. Line L passes through (0, 4) and (10, 1).
Find the equation of line L in the form y = mx + c.
Show solution
- Use the two points (0, 4) and (10, 1).
- Gradient m = change in y⁄change in x = 1 − 4⁄10 − 0 = −3⁄10 = −0.3
- The line crosses the y-axis at (0, 4), so c = 4.
- Substitute into y = mx + c: y = −0.3x + 4
May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q19
Line L has the equation y = 3x + 7.
(a) Write down the gradient of line L.
(b) Point P (4, p) and point W (w, −11) lie on line L. (i) Work out the value of p. (ii) Work out the value of w.
Show solution
- In y = mx + c, the number in front of x is the gradient.
- Gradient = 3
- P is on the line, so substitute x = 4: p = 3 × 4 + 7 = 12 + 7 = 19
- W is on the line, so substitute y = −11: −11 = 3w + 7.
- Subtract 7 from both sides: −18 = 3w.
- Divide by 3: w = −6
Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q21
The line L is shown on the grid. It crosses the y-axis at 5 and goes up 3 for every 1 across. Find the equation of the line L in the form y = mx + c.
Show solution
- The line goes up 3 for every 1 across, so the gradient m = 3.
- It crosses the y-axis at 5, so c = 5.
- Substitute into y = mx + c: y = 3x + 5
Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q26
Line L is drawn on the grid. It crosses the y-axis at 9 and goes up 3 for every 1 across. Find the equation of line L in the form y = mx + c.
Show solution
- The line goes up 3 for every 1 across, so the gradient m = 3.
- It crosses the y-axis at 9, so c = 9.
- Substitute into y = mx + c: y = 3x + 9
0580 Equations of Linear Graphs past paper questions: Extended (Papers 2 and 4)
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q17
Find the equation of the straight line that passes through the points (2, 0) and (0, 4). Give your answer in the form y = mx + c.
Show solution
- Gradient = change in y⁄change in x = 4 − 0⁄0 − 2 = 4⁄−2 = −2, so m = −2.
- The point (0, 4) is on the y-axis, so the y-intercept is c = 4.
- Put these into y = mx + c: y = −2x + 4
- Check: x = 2 gives y = −4 + 4 = 0 ✓
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