What is A8: Coordinates?

A coordinate (x,y) gives horizontal position first and vertical position second. Positive x is right, negative x is left, positive y is up and negative y is down.

How many marks are A8 questions worth?

In verified past-paper appearances, A8 questions on Edexcel GCSE Maths 1MA1 have been worth 2–4 marks.

Which past exam papers has A8 appeared in?

A8 has appeared in 8 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Work with coordinates in all four quadrants

Coordinates

Worked answers, methods and verified real exam appearances for A8 on Edexcel GCSE Maths 1MA1.

Explanation

A coordinate $(x,y)$ gives horizontal position first and vertical position second. Positive $x$ is right, negative $x$ is left, positive $y$ is up and negative $y$ is down.
The signs identify the quadrant, numbered anticlockwise from the top right.
Find a displacement by subtracting starting coordinates from ending coordinates.
Find a midpoint by averaging the two $x$ -coordinates and separately averaging the two $y$ -coordinates.
Examiners expect coordinate order and negative signs to be preserved, with the two component calculations shown clearly.

The four coordinate quadrants and the signs of

(x,y)

in each.

Worked example

Find the midpoint of the segment joining $(-5,2)$ and $(3,-4)$ .

1.Average the $x$ -coordinates: $\dfrac{-5+3}{2}=-1$ .
2.Average the $y$ -coordinates: $\dfrac{2+(-4)}{2}=-1$ .
3.Write the coordinates in $(x,y)$ order.

Answer: $(-1,-1)$ .

Common mistakes

Don't write the vertical coordinate first and swap $(x,y)$ .
Don't lose a negative sign when averaging coordinates.
Don't find half the coordinate differences but forget to add them to an endpoint.

Exam tip

Write the midpoint as two separate averages before combining them into one ordered pair.

Worked practice

Tier 1 · Easy

State the quadrant containing the point

(-4,3)

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Quadrant II	1	The $x$ -coordinate is negative and the $y$ -coordinate is positive, which places the point in quadrant II.

Tier 3 · Hard

The point

P

divides the line segment from

A(-6,5)

B(4,-5)

in the ratio

AP:PB=3:2

. Find the coordinates of

P

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$P=(0,-1)$	3	The vector from $A$ to $B$ is $(10,-10)$ . Since $AP$ is $3/5$ of the whole segment, $AP=(6,-6)$ . Adding this to $A$ gives $P=(-6,5)+(6,-6)=(0,-1)$ .

Tier 1 · Easy

The point

(a,-4)

is in the bottom-left quadrant of the coordinate grid. Write down whether

a

is positive or negative.

(1)

(Total for Question 3 is 1 mark)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$a$ is negative (or $a<0$ )	1	In the bottom-left quadrant, both coordinates are negative. The $x$ -coordinate is $a$ , so $a$ must be negative.

Tier 2 · Standard

A point moves from

(-4,-1)

(2,3)

. Write down its horizontal displacement and its vertical displacement.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$6$ units right $4$ units up	2	For the horizontal displacement, $2-(-4)=6$ , so it moves $6$ units right. For the vertical displacement, $3-(-1)=4$ , so it moves $4$ units up.

Tier 3 · Hard

The midpoint of

A(-7,4)

and

B

M(-2,-1)

. Find the coordinates of

B

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$B=(3,-6)$	3	From $A$ to the midpoint, the change is $(5,-5)$ . The same change takes the midpoint to $B$ , so $B=(-2+5,-1-5)=(3,-6)$ .

Tier 2 · Standard

ABCD

is a rectangle whose vertices are named in order and whose sides are parallel to the coordinate axes. Three vertices are

A(-3,2)

B(5,2)

and

C(5,-4)

. Find the coordinates of

D

and the perimeter of the rectangle.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$D=(-3,-4)$ Perimeter $=28$ units	3	Since $AD$ is vertical, $D$ has the same $x$ -coordinate as $A$ . Since $CD$ is horizontal, $D$ has the same $y$ -coordinate as $C$ , so $D=(-3,-4)$ . The side lengths are $8$ and $6$ , giving perimeter $2(8+6)=28$ units.

Tier 3 · Hard

M

is the midpoint of

P(-6,4)

and

Q(8,-2)

. Point

R

3

units left and

5

units above

M

. Find the coordinates of

R

and write down the quadrant containing

R

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$R=(-2,6)$ The top-left quadrant, where $x$ is negative and $y$ is positive.	3	The midpoint is $M=\left(\dfrac{-6+8}{2},\dfrac{4-2}{2}\right)=(1,1)$ . Moving $3$ units left and $5$ units up gives $R=(1-3,1+5)=(-2,6)$ . A negative $x$ -coordinate and positive $y$ -coordinate place $R$ in the top-left quadrant.

Tier 3 · Hard

The points

A(-5,2)

B(1,7)

C(6,-1)

and

D

occur consecutively around a parallelogram. Find the coordinates of

D

. Check your answer by showing that the two diagonals have the same midpoint.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$D=(0,-6)$ Both diagonal midpoints are $\left(\dfrac12,\dfrac12\right)$	4	For consecutively named vertices, $\overrightarrow{AD}=\overrightarrow{BC}=(5,-8)$ . Hence $D=(-5,2)+(5,-8)=(0,-6)$ . The midpoint of $AC$ is $\left(\dfrac{-5+6}{2},\dfrac{2-1}{2}\right)=\left(\dfrac12,\dfrac12\right)$ , and the midpoint of $BD$ is $\left(\dfrac{1+0}{2},\dfrac{7-6}{2}\right)=\left(\dfrac12,\dfrac12\right)$ . Naming the vertices consecutively fixes this single configuration.

Tier 3 · Hard

The point

P=(x,y)

has a negative

y

-coordinate. The points

A(-4,6)

and

B(1,0)

are fixed. The midpoint of

AP

lies on the

y

-axis and

PB=5

units. Find the coordinates of

P

. Verify both conditions and explain why the point is unique.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$P=(4,-4)$ The midpoint of $AP$ is $(0,1)$ and $PB=5$ The other distance solution is $(4,4)$ , but its $y$ -coordinate is positive, so it is rejected	5	The midpoint of $AP$ has $x$ -coordinate $\dfrac{-4+x}{2}$ . Since it lies on the $y$ -axis, $\dfrac{-4+x}{2}=0$ , so $x=4$ . Now $PB=5$ gives $\sqrt{(4-1)^2+(y-0)^2}=5$ , so $9+y^2=25$ and $y=\pm4$ . The stated negative $y$ -coordinate gives $P=(4,-4)$ . Its midpoint with $A$ is $(0,1)$ and $PB=\sqrt{3^2+(-4)^2}=5$ . The only other candidate, $(4,4)$ , has a positive $y$ -coordinate, so $P$ is unique.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-06	2F	Q8	4	Allowed	Foundation	QP MS
2022-11	3F	Q9	3	Allowed	Foundation	QP MS
2022-11	2H	Q13	2	Allowed	Higher	QP MS
2019-11	1F	Q10	2	Non-calculator	Foundation	QP MS
2024-11	1F	Q8	3	Non-calculator	Foundation	QP MS
2022-11	1F	Q15	2	Non-calculator	Foundation	QP MS
2023-06	1F	Q9	4	Non-calculator	Foundation	QP MS
2022-06	2F	Q8	4	Allowed	Foundation	QP MS

Other points in A Algebra · equations and reasoning

Want help turning this into marks?

Bring A8 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call