IGCSE Mathematics: Coordinate Geometry — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Coordinate Geometry revision notes

Tier note: items marked (Extended) are Extended only in the 0580 syllabus: calculating a gradient from the coordinates of two points, including finding the equation of a line through two given points (E3.3; Core finds a gradient from a grid only), length and midpoint (E3.4) and perpendicular lines (E3.7). All other questions use Core content.

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Questions

1. Points B(4, 1) and C(4, 6) lie on a grid.

(a) State what is true about the line BC (horizontal, vertical, or neither). [1] (b) Point D has the same y-coordinate as C and lies 3 units to the left of C. Write down the coordinates of D. [1]

2. Complete a table of values for y = 2x − 1 for x = −2, 0 and 2, and state two of the points you would plot to draw the graph. [3]

3. (Extended) Find the gradient of the line joining the points (1, 4) and (5, 12). [2]

4. (Extended) A(1, 2) and B(7, 10) are two points.

(a) Calculate the length of AB. [2] (b) Find the coordinates of the midpoint of AB. [2]

5. (Extended) Find the equation of the straight line that passes through (2, 3) and (4, 9), giving your answer in the form y = mx + c. [3]

6. A line has equation y = 3x − 7.

(a) State the gradient of a line that is parallel to this line. [1] (b) Find the equation of the line parallel to y = 3x − 7 that passes through (1, 4). [2]

7. (Extended) A line L has equation y = 4x − 1.

(a) State the gradient of a line perpendicular to L. [1] (b) Find the equation of the line perpendicular to L that passes through (2, 5). [2]

8. State the gradient of a horizontal line, and explain why a vertical line does not have a defined gradient. [2]

9. State whether a line with gradient −3 slopes upward or downward from left to right as x increases, and explain how you know. [2]

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Answers

1. (a) Vertical — both points have the same x-coordinate [1]. (b) (1, 6) [1].

2. x = −2: y = 2(−2) − 1 = −5; x = 0: y = −1; x = 2: y = 2(2) − 1 = 3 [2]. Any two of the points (−2, −5), (0, −1), (2, 3) could be plotted and joined with a straight line [1].

3. Gradient = (12 − 4) / (5 − 1) = 8 / 4 = 2 [2].

4. (a) AB = √((7 − 1)² + (10 − 2)²) [1] = √(36 + 64) = √100 = 10 [1]. (b) Midpoint = ((1 + 7)/2, (2 + 10)/2) = (4, 6) [2].

5. Gradient = (9 − 3) / (4 − 2) = 6/2 = 3 [1]. Using (2, 3): 3 = 3(2) + c → c = 3 − 6 = −3 [1]. y = 3x − 3 [1].

6. (a) 3 — parallel lines have equal gradient [1]. (b) y = 3x + c, using (1, 4): 4 = 3(1) + c → c = 1 [1]. y = 3x + 1 [1].

7. (a) L has gradient 4, so a perpendicular gradient satisfies 4 × m = −1 → m = −¼ [1]. (b) y = −¼x + c, using (2, 5): 5 = −¼(2) + c → c = 5.5 [1]. y = −¼x + 5.5 [1].

8. A horizontal line has gradient 0, since the y-coordinate never changes, so the change in y is always 0 [1]. A vertical line does not have a defined gradient because the x-coordinate never changes between any two points on it, so the gradient formula would require dividing by a change in x of 0, which is undefined [1].

9. The line slopes downward from left to right [1], because a negative gradient means y decreases as x increases, which is what a downward slope, read left to right, looks like [1].

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Where marks are usually lost

• Using rise/run the wrong way round when finding a gradient (dividing the change in x by the change in y).
• Forgetting to square-root at the end of the distance formula, leaving the answer as the sum of squares.
• Adding the two y-coordinates and two x-coordinates for the midpoint but forgetting to divide by 2.
• Using the negative reciprocal rule backwards — multiplying instead of taking the reciprocal, or forgetting the sign flip.
• Substituting a point into y = mx + c to find c, then quoting the wrong final equation (still showing the point’s coordinates instead of m and c).

Examiner report insight

• Not every gradient question needs a negative reciprocal – that rule is for finding a perpendicular line’s gradient. If a question only asks for the equation of the line itself, use the gradient as calculated, without inverting or flipping its sign.

Source: Cambridge International, 0580 Mathematics Principal Examiner Report, June 2024 series, Paper 22 (verified 2026-09-02).

Approaching coordinate geometry questions

Almost every question type in this topic reduces to the same first step: correctly labelling which point is (x1, y1) and which is (x2, y2), then keeping that labelling consistent through every later calculation on the same points. A gradient, length or midpoint calculated with the coordinates swapped partway through is the single most common source of an otherwise fully-understood answer coming out wrong, so it is worth writing the two points down explicitly before substituting into any formula, rather than trying to substitute directly from the question.