Mathematics 0580 · syllabus C3.1, C3.2, C3.3, C3.5, C3.6, E3.1, E3.2, E3.3, E3.4, E3.5, E3.6, E3.7
Cambridge IGCSE Mathematics 0580 — Coordinate geometry topical past papers
Straight-line graphs: plotting lines, gradient, y = mx + c and parallel lines, and on the Extended route length, midpoint and perpendicular lines.
- Questions
- 198
- Years
- 2009–2026
- With mark scheme
- 92%
- Questions 2022–2026
- 74
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What Coordinate geometry covers
Coordinate geometry describes straight lines with algebra. You plot points and draw the line for an equation such as y = 3x − 2, find the gradient as the change in y divided by the change in x, and read and write equations in the form y = mx + c, where m is the gradient and c the y-intercept. Vertical lines have equations x = k and horizontal lines y = k. Parallel lines have the same gradient, so the line parallel to y = 3x − 2 through (2, 1) is y = 3x − 5.
Core candidates find gradients from a grid. Extended candidates also work out a gradient from any two points, as (y₂ − y₁)/(x₂ − x₁), find the length of a segment with Pythagoras' theorem and find its midpoint, and work with equations in other forms, such as 3x + 2y = 12, which need rearranging to show the gradient. Perpendicular lines have gradients that multiply to −1, which gives the equation of the perpendicular bisector of a segment: it passes through the midpoint, with the negative reciprocal of the segment's gradient.
Lines, gradients, y = mx + c and parallel lines are Core content, so they can be set on all four papers; length, midpoint and perpendicular lines are Extended, examined only on Papers 2 and 4. The topic supports graphs of functions, regions defined by inequalities, travel graphs, where the gradient is a speed, and differentiation, which finds the gradient of a curve.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- C3.1 Coordinates
- C3.2 Drawing linear graphs
- C3.3 Gradient of linear graphs
- C3.5 Equations of linear graphs
- C3.6 Parallel lines
- E3.1 Coordinates
- E3.2 Drawing linear graphs
- E3.3 Gradient of linear graphs
- E3.4 Length and midpoint
- E3.5 Equations of linear graphs
- E3.6 Parallel lines
- E3.7 Perpendicular lines
What you should be able to do
- Plot points and draw straight-line graphs from an equation or a table of values
- Find the gradient of a line from a graph, or from two points on the Extended route
- Find and interpret the equation y = mx + c of a line, including x = k and y = k
- Write the equation of a line through a given point, parallel to a given line
- Find the length of a line segment and its midpoint (Extended)
- Find the equation of a perpendicular line or a perpendicular bisector (Extended)
Key ideas
- Gradient m = (y₂ − y₁)/(x₂ − x₁), the change in y divided by the change in x
- y = mx + c has gradient m and crosses the y-axis at (0, c)
- Parallel lines have equal gradients; perpendicular lines have gradients with m₁ × m₂ = −1
- Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2); length = √((x₂ − x₁)² + (y₂ − y₁)²)
How Coordinate geometry has been examined
From 2022 to 2026, Coordinate geometry was on Paper 4 · Calculator (Extended) in 9 of 14 exam series, worth about 5% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 1 · Non-calculator (Core) | 16 | 2% | 11 of 14 |
| Paper 2 · Non-calculator (Extended) | 24 | 4% | 12 of 14 |
| Paper 3 · Calculator (Core) | 18 | 4% | 9 of 14 |
| Paper 4 · Calculator (Extended) | 16 | 5% | 9 of 14 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 4.
Filed in Momentum under: Lines.
A worked example
A is the point (−2, 5) and B is (4, −1). Find the length of AB, its midpoint, and the equation of the perpendicular bisector of AB.
- Length: AB = √((4 − (−2))² + (−1 − 5)²) = √(36 + 36) = √72 = 6√2, which is about 8.49 units.
- Midpoint: ((−2 + 4)/2, (5 + (−1))/2) = (1, 2).
- Gradient of AB = (−1 − 5)/(4 − (−2)) = −6/6 = −1, so the perpendicular gradient is 1.
- The bisector passes through (1, 2) with gradient 1: 2 = 1 + c, so c = 1 and the equation is y = x + 1.
- Check: (0, 1) lies on the bisector, and its distance from A and from B is √20 in both cases.
Answer: AB = 6√2 ≈ 8.49; midpoint (1, 2); perpendicular bisector y = x + 1
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Reading the gradient of 3x + 2y = 12 as 3: rearrange to y = −1.5x + 6 first, so the gradient is −1.5.
- Putting the change in x on top when finding a gradient: the gradient is the change in y divided by the change in x.
- Using the same gradient for a perpendicular line: perpendicular gradients are negative reciprocals, so a line perpendicular to one with gradient 2/3 has gradient −3/2.
- Getting the sign of a gradient wrong: a line that falls from left to right has a negative gradient.
- Leaving an equation unsimplified: lines are expected in a fully simplified form, such as y = 2x + 1 rather than y − 5 = 2(x − 2).
Exam technique
- Find m first, then substitute a point into y = mx + c to find c, and write out the full equation.
- Sketch the points before calculating; a rough picture shows whether a gradient should be positive or negative.
- For a perpendicular bisector, find the midpoint and the gradient of the segment first, then use the negative reciprocal.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.