What is G11: Coordinate geometry?

Coordinate geometry turns a diagram into exact calculations. The midpoint of (x_1,y_1) and (x_2,y_2) is (x_1+x_2/2,y_1+y_2/2).

How many marks are G11 questions worth?

In verified past-paper appearances, G11 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has G11 appeared in?

G11 has appeared in 10 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G11

Solve geometrical problems on coordinate axes

Coordinate geometry

Worked answers, methods and verified real exam appearances for G11 on Edexcel GCSE Maths 1MA1.

Explanation

Coordinate geometry turns a diagram into exact calculations. The midpoint of $(x_1,y_1)$ and $(x_2,y_2)$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .
Horizontal and vertical coordinate changes form a right triangle, so distance is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .
Coordinate differences can prove equal side lengths and identify horizontal or vertical lines; areas can be found from base and perpendicular height or by enclosing and subtracting simpler shapes.
Keep brackets when subtracting negative coordinates and show the substitution.
The axes and calculated differences, not the sketch's appearance, supply the evidence the examiner needs.

Coordinate differences form a right triangle for gradient and distance.

Worked example

Points $A(-3,2)$ and $B(5,8)$ are endpoints of a segment. Find its midpoint and exact length.

1.Midpoint $=\left(\frac{-3+5}{2},\frac{2+8}{2}\right)=(1,5)$ .
2.Coordinate changes are $5-(-3)=8$ and $8-2=6$ .
3. $AB=\sqrt{8^2+6^2}=\sqrt{100}=10$ .

Answer: The midpoint is $(1,5)$ and $AB=10$ .

Common mistakes

Don't calculate the midpoint by subtracting coordinates instead of averaging each coordinate pair.
Don't write $5-(-3)=2$ , losing the second negative sign.

Exam tip

In a 'show that' coordinate proof, display the gradients or squared lengths that establish the claimed property.

Worked practice

Tier 1 · Easy

Find the midpoint of the line segment joining

(-5,2)

(7,8)

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$(1,5)$	2	Average the coordinates: $x=(-5+7)/2=1$ and $y=(2+8)/2=5$ . The midpoint is $(1,5)$ .

Tier 2 · Standard

Points

A(-1,4)

and

B(5,-4)

are joined. Find the exact length

AB

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$10$	3	The coordinate changes are $5-(-1)=6$ and $-4-4=-8$ . Therefore $AB=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10$ .

Tier 3 · Hard

Triangle

ABC

has vertices

A(-2,1)

B(4,3)

and

C(2,9)

. Show that the triangle is right-angled and isosceles, then work out its area.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Right-angled at $B$ $AB=BC=2\sqrt{10}$ Area $=20$ square units	5	The gradients are $m_{AB}=2/6=1/3$ and $m_{BC}=6/(-2)=-3$ , whose product is $-1$ , so the angle at $B$ is $90^\circ$ . Also $AB=\sqrt{6^2+2^2}=\sqrt{40}=2\sqrt{10}$ and $BC=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}$ , so the triangle is isosceles. Its perpendicular equal sides have area $\frac12(2\sqrt{10})^2=20$ square units.

Tier 1 · Easy

Points

P(-6,4)

and

Q(3,4)

are joined. Work out the length

PQ

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$9$ units	1	The points have the same $y$ -coordinate, so the segment is horizontal. Its length is $3-(-6)=9$ .

Tier 2 · Standard

The midpoint of

AB

(2,1)

. Point

A

(-3,4)

. Find the coordinates of

B

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$B=(7,-2)$	2	Let $B=(x,y)$ . For the $x$ -coordinate, $(-3+x)/2=2$ , so $x=7$ . For the $y$ -coordinate, $(4+y)/2=1$ , so $y=-2$ . Therefore $B=(7,-2)$ .

Tier 3 · Hard

The points

A(1,2)

B(7,2)

C(9,7)

and

D(3,7)

are joined in order. Show that

ABCD

is a parallelogram and work out its area.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$ABCD$ is a parallelogram; area $=30$ square units	4	$AB$ and $DC$ are both horizontal and each has length $6$ . From $A$ to $D$ and from $B$ to $C$ , the coordinate changes are both $2$ right and $5$ up, so those sides are equal and parallel. Hence $ABCD$ is a parallelogram. Taking $AB=6$ as the base, the perpendicular height is $7-2=5$ , so the area is $6\times5=30$ square units.

Tier 2 · Standard

Triangle

ABC

has vertices

A(-4,-2)

B(8,-2)

and

C(3,5)

. Work out the area of the triangle.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$42$ square units	3	$AB$ is horizontal with length $8-(-4)=12$ . The perpendicular height from $C$ to line $AB$ is $5-(-2)=7$ . Therefore the area is $\frac12\times12\times7=42$ square units.

Tier 3 · Hard

Point

P

lies on the

y

-axis and is equidistant from

A(-3,1)

and

B(5,5)

. Find the coordinates of

P

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$P=(0,5)$	4	Let $P=(0,y)$ . Equating squared distances gives $3^2+(y-1)^2=5^2+(y-5)^2$ . Expanding and cancelling $y^2$ gives $10-2y=50-10y$ , so $8y=40$ and $y=5$ . Therefore $P=(0,5)$ ; indeed, both distances are $5$ .

Tier 3 · Hard

Points

A(-4,1)

and

B(8,7)

are endpoints of a diameter of a circle. Work out the centre and the exact radius of the circle. Point

C

has coordinates

(5,-2)

. Show that

C

lies on the circle.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Centre $(2,4)$ ; radius $3\sqrt5$ units; $C$ lies on the circle	4	The centre is the midpoint of $AB$ , namely $(({-4+8})/2,(1+7)/2)=(2,4)$ . Its squared distance to $A$ is $(-6)^2+(-3)^2=45$ , so the radius is $\sqrt{45}=3\sqrt5$ . From the centre to $C$ the change is $(3,-6)$ , whose squared length is $3^2+(-6)^2=45$ . Therefore $C$ is exactly one radius from the centre and lies on the circle.

Q10

Tier 3 · Hard

Points

A(-4,0)

B(1,6)

and

C(6,0)

are consecutive vertices of parallelogram

ABCD

. Work out the coordinates of

D

. Show that the parallelogram is a rhombus but not a square.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$D=(1,-6)$ ; $ABCD$ is a rhombus but not a square	5	The diagonals of a parallelogram share a midpoint. Midpoint $AC$ is $(1,0)$ , so this must also be the midpoint of $BD$ ; with $B=(1,6)$ , this gives $D=(1,-6)$ . The four squared side lengths are $AB^2=BC^2=CD^2=DA^2=5^2+6^2=61$ , so all four sides are equal and the shape is a rhombus. Its diagonals have lengths $AC=10$ and $BD=12$ , so they are unequal; a square has equal diagonals, hence this is not a square. The midpoint condition makes $D$ unique.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2021-11	2H	Q11	5	Allowed	Higher	QP MS
2024-06	1H	Q23	5	Non-calculator	Higher	QP MS
2023-11	2H	Q12	3	Allowed	Higher	QP MS
2024-11	2H	Q21	5	Allowed	Higher	QP MS
2019-11	2H	Q25	5	Allowed	Higher	QP MS
2022-06	2H	Q5	4	Allowed	Higher	QP MS
2022-11	1F	Q15	2	Non-calculator	Foundation	QP MS
2022-06	2F	Q25	4	Allowed	Foundation	QP MS
2024-11	3H	Q16	5	Allowed	Higher	QP MS
2021-11	3H	Q13	5	Allowed	Higher	QP MS

Other points in G Geometry and measures

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