Coordinates in four quadrants - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 18 exam-style questions on Coordinates in four quadrants for Edexcel GCSE Maths (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.

18 questions
1–5 marks each
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18 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 \((-6.4, 3.8)\).
\(M\) is the point \((1.2, -0.5)\).

\(M\) is the midpoint of the line segment \(AB\).

Find the coordinates of \(B\).

Model answer

From \(A\) to \(M\): x goes up by \(1.2 - (-6.4) = 7.6\), y goes down by \(3.8 - (-0.5) = 4.3\)
From \(M\) to \(B\) is the same step again:
x: \(1.2 + 7.6 = 8.8\)
y: \(-0.5 - 4.3 = -4.8\)
\(B = (8.8, -4.8)\)

Explanation

\(M\) is halfway from \(A\) to \(B\), so the step from \(M\) to \(B\) is the same as the step from \(A\) to \(M\). Equivalently, \(B = 2M - A\) for each coordinate. A common mistake is to find the midpoint of \(A\) and \(M\) instead.

Mark scheme

1 mark
M1 for a correct method for one coordinate, e.g. \(2 \times 1.2 - (-6.4)\) or \(1.2 + 7.6\), or \(2 \times (-0.5) - 3.8\) or \(-0.5 - 4.3\)
oe, e.g. \(\dfrac{-6.4 + x}{2} = 1.2\); correct answer with no working scores full marks
1 mark
A1 for (8.8, −4.8)
cao

Question 2

2 marks

\(B\) is the point \((-5, -2)\) and \(C\) is the point \((3, -2)\).

Triangle \(ABC\) is isosceles with \(AB = AC\).
The point \(A\) lies on the line \(y = 4\).

Find the coordinates of \(A\).

Model answer

As \(AB = AC\), \(A\) is directly above the midpoint of \(BC\) (\(BC\) is horizontal).
Midpoint of \(BC\): \(x = \dfrac{-5 + 3}{2} = -1\)
\(A\) is on \(y = 4\), so \(A = (-1, 4)\)

Explanation

Every point the same distance from \(B\) and \(C\) lies on the perpendicular bisector of \(BC\). Since \(BC\) is horizontal, that is the vertical line through its midpoint, \(x = -1\). Where this meets \(y = 4\) is \(A\).

Mark scheme

1 mark
M1 for finding the x-coordinate of the midpoint of \(BC\), \(x = -1\)
oe, e.g. \(\dfrac{-5 + 3}{2}\); correct answer with no working scores full marks
1 mark
A1 for (−1, 4)
cao

Question 3

3 marks

The grid shows the points \(A\) and \(B\).

\(ABCD\) is a square.
The vertices \(C\) and \(D\) are above the line \(AB\).

Find the coordinates of \(C\) and the coordinates of \(D\).

A coordinate grid with x from −5 to 5 and y from −3 to 7; each grid square is 1 unit across, and the axes cross at the origin O. Point A is marked with a cross at (−2, −1). Point B is marked with a cross at (3, 1). A straight line segment joins A (−2, −1) to B (3, 1).

Model answer

\(A = (-2, -1)\), \(B = (3, 1)\)
From \(A\) to \(B\): 5 right, 2 up.
A side at right angles to \(AB\), going upwards: 2 left, 5 up.
\(C = (3 - 2, 1 + 5) = (1, 6)\)
\(D = (-2 - 2, -1 + 5) = (-4, 4)\)

Explanation

All sides of a square are equal and meet at right angles. Turning the step '5 right, 2 up' through 90° anticlockwise gives '2 left, 5 up', which is the same length and perpendicular (gradients \(\frac{2}{5}\) and \(-\frac{5}{2}\) multiply to \(-1\)). Add this step to \(B\) to get \(C\) and to \(A\) to get \(D\).

Mark scheme

1 mark
M1 for the step from \(A\) to \(B\) (5 right, 2 up) turned through 90° to give a step of 2 left and 5 up
oe, e.g. one of \(C\) or \(D\) correct; correct answer with no working scores full marks
1 mark
A1 for \(C = (1, 6)\)
cao
1 mark
A1 for \(D = (-4, 4)\)
cao

Question 4

4 marks

\(A\) is the point \((-2, 5)\) and \(B\) is the point \((6, 3)\).

The point \(P\) lies on the x-axis.
\(P\) is the same distance from \(A\) as it is from \(B\).

Find the coordinates of \(P\).

Model answer

Let \(P = (x, 0)\)
\(PA^2 = (x + 2)^2 + 5^2 = x^2 + 4x + 29\)
\(PB^2 = (x - 6)^2 + 3^2 = x^2 - 12x + 45\)
\(x^2 + 4x + 29 = x^2 - 12x + 45\)
\(16x = 16\)
\(x = 1\)
\(P = (1, 0)\)

Explanation

Write \(P\) as \((x, 0)\) because it is on the x-axis, then use Pythagoras' theorem for both distances. Setting the squared distances equal avoids square roots, and the \(x^2\) terms cancel. Check: \(PA = PB = \sqrt{34}\).

Mark scheme

1 mark
M1 for \(P = (x, 0)\) and \(PA^2 = (x + 2)^2 + 5^2\)
oe; correct answer with no working scores full marks
1 mark
M1 for \(PB^2 = (x - 6)^2 + 3^2\) and setting \(PA^2 = PB^2\)
oe
1 mark
M1 for expanding and simplifying to a linear equation, e.g. \(16x = 16\)
oe; or a correct method using the perpendicular bisector of \(AB\), \(y = 4x - 4\)
1 mark
A1 for (1, 0)
cao

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