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GCSE Maths · Geometry & measures

Coordinate geometry GCSE Questions and Worked Answers

Coordinates locate points using horizontal x then vertical y. Compare the coordinates to find horizontal and vertical changes; these give midpoint, distance and gradient. Use the scale on each axis, not a visual estimate.

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Start with the meaning

What you need to know about coordinate geometry

On a grid, two numbers give a point's address. Start where the axes cross, called the origin (0, 0). For the point (3, 2), go 3 units right and 2 units up. The first number, x, records horizontal position; the second, y, records vertical position. Negative x means left, and negative y means down.

See the idea first

One journey supplies three useful measurements

From A = (1, 1) to B = (7, 5), the horizontal change is 7 − 1 = 6 and the vertical change is 5 − 1 = 4. Halfway along the straight journey means half of each change, reaching (4, 3). We call this point the midpoint. The same two changes are the legs of a right triangle whose sloping side is AB.

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From A to B, move 6 right and 4 up. Half of each move reaches M = (4,3). The dashed right triangle also lets you calculate AB using Pythagoras.
From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Find a midpoint

What the problem asks: Find the midpoint of (2, 3) and (8, 7).

How to solve it: Average the two horizontal positions: (2 + 8)/2 = 5. Average the vertical positions: (3 + 7)/2 = 5. The midpoint is (5, 5).

Find a straight-line distance

What the problem asks: Find the distance between (1, 2) and (4, 6).

How to solve it: The changes are 3 across and 4 up. Pythagoras gives length √(3² + 4²) = 5 units; adding 3 + 4 measures a bent route instead.

Find gradient or a line equation

What the problem asks: Find the gradient and equation of the line through (1, 2) and (4, 8).

How to solve it: Gradient means vertical change per one horizontal unit. Here it is (8 − 2)/(4 − 1) = 6/3 = 2. For the equation, write y = 2x + c. Substitute (1, 2): 2 = 2 + c, so c = 0 and y = 2x.

A reliable routine

Use coordinates to solve a geometry problem

Use coordinate differences when a problem asks about a segment's halfway point, length or slope. The method depends on the requested quantity: averages locate the midpoint, Pythagoras measures the diagonal, and rise divided by run gives gradient.

  1. Identify what is required and write both points in the same x, y order.
  2. Work out horizontal and vertical changes, subtracting in the same direction.
  3. Use averages for midpoint, Pythagoras for distance, or vertical change ÷ horizontal change for gradient.
  4. Check the result against the positions and include units if the axes have them.

Check: Vertical lines have zero horizontal change, so their gradient is undefined. Their equation is x = a constant, not y = mx + c.

Fully worked

Coordinate geometry GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Midpoint across the axes

2 marks
Question

Find the midpoint of A = (−3, 2) and B = (5, 8).

xM=−3+52=1x_M=\frac{-3+5}{2}=1 yM=2+82=5y_M=\frac{2+8}{2}=5

The midpoint is (1, 5). It lies halfway in both coordinate directions.

Example 2

A missing endpoint

3 marks
Question

A = (2, −1). The midpoint of AB is M = (5, 3). Find B.

From A to M, move 3 across and 4 up: a displacement written (3, 4), not the coordinates of a new point. Repeat that same change from M to B.

B=(5+3, 3+4)=(8,7)B=(5+3,\ 3+4)=(8,7)

Check: averaging (2, −1) and (8, 7) gives (5, 3).

Example 3

Distance

3 marks
Question

Find the distance between (−2, 1) and (4, 9).

horizontal change=4−(−2)=6\text{horizontal change}=4-(-2)=6 vertical change=9−1=8\text{vertical change}=9-1=8 d2=62+82=100d^2=6^2+8^2=100 d=10 unitsd=10\text{ units}

Length is positive.

Example 4

A negative gradient

2 marks
Question

Find the gradient through (1, 7) and (5, −1).

m=−1−75−1m=\frac{-1-7}{5-1} =−84=−2=\frac{-8}{4}=-2

Moving one unit right lowers the line by 2 units.

Example 5

Line through two points

4 marks
Question

Find the equation of the line through (2, 5) and (6, 13).

m=13−56−2=2m=\frac{13-5}{6-2}=2

Write y = 2x + c and substitute (2, 5):

5=2(2)+c5=2(2)+c c=1c=1

Thus y = 2x + 1. Check (6, 13): 2 × 6 + 1 = 13.

Example 6

Perpendicular line

Higher only4 marks
Question

Find the equation of the line perpendicular to y = 2x + 1 through (4, 3).

Non-vertical perpendicular gradients multiply to −1, so the new gradient is −1/2.

y=−12x+cy=-\frac12x+c 3=−12(4)+c3=-\frac12(4)+c c=5c=5

Hence y=−12x+5y=-\frac12x+5.

10 original questions · total 23 marks

Coordinate geometry GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 28 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed
1

Locate a point

1 mark

Describe how to reach (−4, −2) from the origin.

Show worked answer

Move 4 units left, then 2 units down. Both coordinates are negative.

2

Midpoint

2 marks

Find the midpoint of (1, 4) and (7, 10).

Show worked answer
M=(1+72,4+102)=(4,7)M=\left(\frac{1+7}{2},\frac{4+10}{2}\right)=(4,7)
3

Negative coordinates

2 marks

Find the midpoint of (−6, −3) and (2, 5).

Show worked answer
M=(−6+22,−3+52)=(−2,1)M=\left(\frac{-6+2}{2},\frac{-3+5}{2}\right)=(-2,1)
4

Recover an endpoint

3 marks

A = (−1, 2) and midpoint M = (2, 6). Find B.

Show worked answer

The change A to M is (3, 4). Repeat it:

B=(2+3,6+4)=(5,10)B=(2+3,6+4)=(5,10)
5

Length

3 marks

Find the distance from (1, 1) to (4, 5).

Show worked answer

Changes are 3 and 4.

d=32+42=5 unitsd=\sqrt{3^2+4^2}=5\text{ units}
6

Exact length

3 marks

Find the exact distance from (−1, 2) to (3, 4).

Show worked answer

Changes are 4 and 2.

d=42+22=20=25 unitsd=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt5\text{ units}
7

Gradient

2 marks

Find the gradient through (−2, 5) and (4, 2).

Show worked answer
m=2−54−(−2)=−36=−12m=\frac{2-5}{4-(-2)}=\frac{-3}{6}=-\frac12
8

Vertical line

2 marks

State the equation and gradient of the line through (3, −2) and (3, 7).

Show worked answer

Every point has x = 3, so its equation is x = 3. The horizontal change is zero, so the gradient is undefined.

9

Equation

3 marks

Find the equation of the line through (0, −3) and (2, 5).

Show worked answer
m=5−(−3)2−0=4m=\frac{5-(-3)}{2-0}=4

The y-intercept is −3, so y = 4x − 3.

10

Perpendicular gradient

Higher only2 marks

Find the gradient perpendicular to a line with gradient −3/4.

Show worked answer

The gradients multiply to −1:

m=43m=\frac43

Check: (−3/4)(4/3)=−1(-3/4)(4/3)=-1.

Examiner-style feedback

Common coordinate geometry mistakes

Switching coordinates

Write horizontal x first, vertical y second.

Changing subtraction order halfway

If you subtract B − A for y, do B − A for x too.

Adding changes for distance

The straight-line distance is the hypotenuse, not the sum of the horizontal and vertical legs.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

  1. x across, y up.
  2. Midpoint averages coordinates.
  3. Distance uses Pythagoras.
  4. Gradient is vertical change ÷ horizontal change.
Quick answers

Coordinate geometry FAQ

Can a midpoint have fractional coordinates?

Yes. Halfway between integer grid points need not be another integer grid point.

Is a negative gradient a negative length?

No. Gradient describes direction of change. Distance measures length and cannot be negative.

Build connected skills

What to revise next

Browse all GCSE Maths study guidesReview the worked examplesTry the questions again
Content standards

Curriculum and rights review

A8/G11 coordinates and geometry across tiers, with A9 perpendicular-gradient extension labelled Higher. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references