What is G24: Vectors as translations?

A translation moves every point by the same horizontal and vertical displacement, preserving lengths, angles and orientation. Describe it with a column vector x/y: a positive top component moves right and a negative one moves left; a positive bottom component moves up and a negative one moves down.

How many marks are G24 questions worth?

In verified past-paper appearances, G24 questions on Edexcel GCSE Maths 1MA1 have been worth 2–3 marks.

Which past exam papers has G24 appeared in?

G24 has appeared in 6 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G24

Describe translations as 2D vectors

Vectors as translations

Worked answers, methods and verified real exam appearances for G24 on Edexcel GCSE Maths 1MA1.

Explanation

A translation moves every point by the same horizontal and vertical displacement, preserving lengths, angles and orientation. Describe it with a column vector $\binom{x}{y}$ : a positive top component moves right and a negative one moves left; a positive bottom component moves up and a negative one moves down.
To recover the vector from object and image coordinates, calculate image minus object in each coordinate.
Apply the same vector to every vertex when constructing an image.
Examiners expect a vector, not a destination coordinate, and the object-to-image direction matters.
Matching one pair of corresponding points is enough if the shapes are genuinely translations.

A translation gives every corresponding point the same displacement vector.

Worked example

Point $A(-2,5)$ is translated to $A'(4,1)$ . Find the translation vector.

1.Horizontal displacement $=4-(-2)=6$ .
2.Vertical displacement $=1-5=-4$ .
3.Write horizontal above vertical in a column vector.

Answer: $\binom{6}{-4}$ .

Common mistakes

Don't subtract object minus image and obtain the reverse vector.
Don't write $(4,1)$ , the image coordinate, instead of the displacement vector.

Exam tip

Use image minus object for both coordinates, then check the signs against the visible direction of movement.

Worked practice

Tier 1 · Easy

Point

A=(-3,4)

is translated to

A'=(2,-1)

. Describe the translation as a vector.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$\binom{5}{-5}$	2	Calculate image minus object in each coordinate: $2-(-3)=5$ and $-1-4=-5$ . The translation vector is $\binom{5}{-5}$ .

Tier 2 · Standard

Triangle

ABC

has vertices

A=(1,-2)

B=(5,1)

and

C=(2,4)

. It is translated by

\binom{-4}{3}

. Write the coordinates of the three image vertices.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$A'=(-3,1)$ $B'=(1,4)$ $C'=(-2,7)$	3	Add $-4$ to every $x$ -coordinate and $3$ to every $y$ -coordinate. This gives $A'=(-3,1)$ , $B'=(1,4)$ and $C'=(-2,7)$ .

Tier 3 · Hard

A translation maps

P=(-6,4)

P'=(1,-2)

. The same translation maps

Q=(3,k)

Q'=(10,5)

. Find

k

and describe the translation as a vector.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$k=11$ $\binom{7}{-6}$	4	From $P$ to $P'$ , the displacement is $1-(-6)=7$ horizontally and $-2-4=-6$ vertically, so the vector is $\binom{7}{-6}$ . Therefore the image of $Q=(3,k)$ has $y$ -coordinate $k-6$ . Since $k-6=5$ , $k=11$ .

Tier 1 · Easy

Point

P=(4,-1)

is translated by

\binom{-3}{5}

. Write down the coordinates of the image of

P

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$P'=(1,4)$	1	Add the vector components to the coordinates: $(4-3,-1+5)=(1,4)$ . Therefore $P'=(1,4)$ .

Tier 2 · Standard

A translation by

\binom{6}{-4}

maps point

Q

Q'=(2,7)

. Work out the coordinates of

Q

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$Q=(-4,11)$	2	Reverse the translation: subtract $6$ from the image's $x$ -coordinate and add $4$ to its $y$ -coordinate. Therefore $Q=(2-6,7+4)=(-4,11)$ .

Tier 3 · Hard

A translation by vector

\mathbf{a}

maps

P=(2,-1)

(5,3)

. A second translation by vector

\mathbf{b}

maps

(5,3)

(1,7)

. Work out

\mathbf{a}

, work out

\mathbf{b}

, and write down the single vector that maps

P

directly to

(1,7)

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$\mathbf{a}=\binom{3}{4}$ $\mathbf{b}=\binom{-4}{4}$ Single vector $\binom{-1}{8}$	3	$\mathbf{a}=\binom{5-2}{3-(-1)}=\binom{3}{4}$ and $\mathbf{b}=\binom{1-5}{7-3}=\binom{-4}{4}$ . Successive translations add: the single vector is $\mathbf{a}+\mathbf{b}=\binom{3-4}{4+4}=\binom{-1}{8}$ , and indeed $P=(2,-1)$ maps directly to $(2-1,\,-1+8)=(1,7)$ .

Tier 2 · Standard

Triangle

T

has vertices

(-5,1)

(-2,1)

and

(-4,4)

. After a translation, its image has vertices

(1,-6)

(4,-6)

and

(2,-3)

in the corresponding order. Describe the translation as a vector.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$\binom{6}{-7}$	3	Using any corresponding pair gives image minus object. For example, $(1-(-5),-6-1)=(6,-7)$ . The other pairs give the same displacement, so the translation vector is $\binom{6}{-7}$ .

Tier 3 · Hard

A mapping sends

A=(1,-2)

A'=(-3,4)

B=(5,0)

B'=(1,6)

and

C=(2,3)

C'=(-2,8)

. State whether this mapping can be a translation. Give a reason.

(3)

(Total for Question 8 is 3 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	No, $A$ and $B$ move by $\binom{-4}{6}$ but $C$ moves by $\binom{-4}{5}$ , so the displacement is not the same for every point.	3	The displacements from $A$ to $A'$ and from $B$ to $B'$ are both $\binom{-4}{6}$ . The displacement from $C$ to $C'$ is $\binom{-4}{5}$ . A translation must move every point by the same vector, so this mapping cannot be a translation.

Tier 3 · Hard

Segment

AB

has endpoints

A=(-4,7)

and

B=(8,-1)

. After a translation, the midpoint of its image

A'B'

(9,3)

. Work out the translation vector and the coordinates of

A'

and

B'

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Translation vector $\binom{7}{0}$ $A'=(3,7)$ and $B'=(15,-1)$	4	The midpoint of $AB$ is $\left(\frac{-4+8}{2},\frac{7+(-1)}{2}\right)=(2,3)$ . Its displacement to $(9,3)$ is $\binom{7}{0}$ , which is the translation vector. Adding this vector gives $A'=(3,7)$ and $B'=(15,-1)$ .

Q10

Tier 3 · Hard

Higher only: The line

y=2x+1

is translated by

\binom{3}{k}

. Its image passes through the point

(5,7)

. Work out

k

and write the equation of the image line.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$k=2$ $y=2x-3$	4	The point $(5,7)$ on the image came from $(5-3,7-k)=(2,7-k)$ on the original line. Since $y=2x+1$ there, $7-k=2(2)+1=5$ , so $k=2$ . Under the translation, $x'=x+3$ and $y'=y+2$ . Thus $y'-2=2(x'-3)+1$ , which simplifies to $y'=2x'-3$ . Therefore the image line is $y=2x-3$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-06	1H	Q11	3	Non-calculator	Higher	QP MS
2024-06	1F	Q19	2	Non-calculator	Foundation	QP MS
2022-06	2F	Q22	2	Allowed	Foundation	QP MS
2022-06	2H	Q2	2	Allowed	Higher	QP MS
2019-06	1F	Q26	3	Non-calculator	Foundation	QP MS
2019-06	1H	Q5	3	Non-calculator	Higher	QP MS

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