Geometry problems on coordinate axes - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Geometry problems on coordinate axes for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
Instant marking
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A is the point (1, 2) and B is the point (7, 10).

Work out the length of the line segment AB.

Model answer

Horizontal distance \(= 7 - 1 = 6\), vertical distance \(= 10 - 2 = 8\)
\(AB^2 = 6^2 + 8^2 = 36 + 64 = 100\)
\(AB = 10\)

Explanation

Draw a right-angled triangle with AB as the hypotenuse: the other sides are the differences in the x-coordinates and in the y-coordinates. Then use Pythagoras' theorem.

Mark scheme

1 mark
M1 for \(6^2 + 8^2\) or \(\sqrt{(7 - 1)^2 + (10 - 2)^2}\)
correct answer with no working scores full marks
1 mark
A1 for 10
cao

Question 2

2 marks

A triangle has vertices P(0, 0), Q(4, 0) and R(2, 3).
Ali says, "Triangle PQR is equilateral."

Is Ali correct? You must show how you get your answer.

Model answer

PQ \(= 4 - 0 = 4\)
PR \(= \sqrt{2^2 + 3^2} = \sqrt{13}\)
\(\sqrt{13}\) is not 4 (since \(4 = \sqrt{16}\)), so the sides are not all equal.
Ali is not correct (the triangle is isosceles, not equilateral).

Explanation

An equilateral triangle has three equal sides, so compare the lengths. Comparing squared lengths (13 and 16) avoids square roots altogether.

Mark scheme

1 mark
M1 for finding PR or QR: \(\sqrt{2^2 + 3^2} = \sqrt{13}\)
1 mark
C1 for 'No' with a reason, e.g. PQ = 4 but PR = \(\sqrt{13}\), and \(\sqrt{13}\) ≠ 4
accept \(PR^2 = 13\) and \(PQ^2 = 16\) compared

Question 3

3 marks

A is the point (4, 1).

Find the coordinates of the two points on the y-axis that are 5 units from A.

Model answer

A point on the y-axis is (0, y).
Distance from A: \(4^2 + (y - 1)^2 = 5^2\)
\((y - 1)^2 = 25 - 16 = 9\)
\(y - 1 = 3\) or \(y - 1 = -3\)
\(y = 4\) or \(y = -2\)
The points are (0, 4) and (0, −2).

Explanation

Every point on the y-axis has x-coordinate 0, so its horizontal distance from A is 4. With hypotenuse 5, the vertical distance must be 3, either above or below A - so there are two answers.

Mark scheme

1 mark
M1 for \(4^2 + (y - 1)^2 = 5^2\) or \((y - 1)^2 = 9\)
oe, e.g. a right-angled triangle with sides 4 and 3 and hypotenuse 5; correct answer with no working scores full marks
1 mark
A1 for one correct point: (0, 4) or (0, −2)
1 mark
A1 for both (0, 4) and (0, −2)
cao

Question 4

4 marks

A is the point (1, 3) and B is the point (5, 1).
The point P lies on the x-axis and is the same distance from A as from B.

Find the coordinates of P.

Model answer

Let P = (x, 0).
\(PA^2 = (x - 1)^2 + 3^2\) and \(PB^2 = (x - 5)^2 + 1^2\)
\((x - 1)^2 + 9 = (x - 5)^2 + 1\)
\(x^2 - 2x + 1 + 9 = x^2 - 10x + 25 + 1\)
\(-2x + 10 = -10x + 26\)
\(8x = 16\)
\(x = 2\)
P = (2, 0)

Explanation

Points that are the same distance from A and B lie on the perpendicular bisector of AB. You can either set the two squared distances equal (the \(x^2\) terms cancel) or find the perpendicular bisector, \(y = 2x - 4\), and see where it crosses the x-axis.

Mark scheme

1 mark
M1 for \((x - 1)^2 + 3^2 = (x - 5)^2 + 1^2\)
or for the midpoint (3, 2) and gradient of AB \(= -\frac{1}{2}\); correct answer with no working scores full marks
1 mark
M1 for expanding correctly, e.g. \(x^2 - 2x + 10 = x^2 - 10x + 26\)
or the perpendicular bisector \(y - 2 = 2(x - 3)\)
1 mark
M1 for \(8x = 16\)
or substituting \(y = 0\) into their perpendicular bisector
1 mark
A1 for (2, 0)
cao

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