Maths · 8 Transformations and vectors

8.1 Transformations

Before you start

Plotting and reading coordinates in all four quadrants, including negative coordinates.
Recognising the lines $x = a$ (vertical), $y = b$ (horizontal), $y = x$ and $y = - x$ on a grid.
Finding the midpoint of a line segment (4.3).
Adding, subtracting and multiplying with negative numbers and simple fractions.
Clockwise and anticlockwise, and that a quarter turn is $90 °$ and a half turn is $180 °$ .

Syllabus objectives covered (4)

Recognise, describe and draw the following transformations: Reflection of a shape in a straight line.
Recognise, describe and draw the following transformations: Rotation of a shape about a centre through multiples of 90°.
Recognise, describe and draw the following transformations: Enlargement of a shape from a centre by a scale factor.
Recognise, describe and draw the following transformations: Translation of a shape by a vector \binom{x}{y}.

Part 1 of 8

Object, image, and what each transformation needs

A transformation moves a shape to a new position. The shape you start with is the object. The shape you end with is the image. If the object has vertices $A$ , $B$ and $C$ , label the matching image vertices $A^{'}$ , $B^{'}$ and $C^{'}$ . Then you always know which point went where.
This subtopic has four transformations. A translation slides the shape. A reflection flips it over a straight line. A rotation turns it about a fixed point. An enlargement changes its size from a fixed point.
Translations, reflections and rotations do not change any length or any angle. The image is congruent to the object: the same shape and the same size. An enlargement changes the lengths, so its image is the same shape but, in general, a different size.
So the first question about an object and its image is: are they the same size? If not, it is an enlargement. If they are, it is a translation, a reflection or a rotation.
To tell those three apart, look at how the image sits. If it faces the same way as the object and has only slid, it is a translation. If it has turned, it is a rotation. If it is a mirror image, it is a reflection.
A test for a mirror image: go round the object $A \to B \to C$ and note whether that is clockwise or anticlockwise. Then go round the image $A^{'} \to B^{'} \to C^{'}$ . A reflection reverses the direction. A translation, a rotation and an enlargement keep it.
Each transformation is fixed by different information, and Describe fully means you give all of it. Translation: the column vector. Reflection: the equation of the mirror line. Rotation: the centre, the angle and the direction. Enlargement: the centre and the scale factor.
Name one transformation only. A question about the single transformation wants one name. 'A reflection and then a translation' is two transformations, so it does not answer the question, even if it gets the shape to the right place.
When you draw an image, plot the vertices first, then join them. The syllabus says a ruler must be used for all straight edges, so rule every side.

Part 2 of 8

Translation by a column vector

A translation is given by a column vector $(\binom{x}{y})$ . The top number is the move across: positive to the right, negative to the left. The bottom number is the move up or down: positive up, negative down.
So $(\binom{3}{- 5})$ means 3 squares right and 5 squares down. $(\binom{- 2}{0})$ means 2 squares left and no move up or down.
In a translation every point of the shape moves by the same vector. That is why the image is the same size, faces the same way, and is not turned.
To translate a point, add the top number to the $x$ -coordinate and the bottom number to the $y$ -coordinate.
$(\binom{a}{b}) : (x, y) \to (x + a, y + b)$
Example: translate the triangle with vertices $(1, 1)$ , $(4, 1)$ and $(1, 3)$ by $(\binom{3}{- 5})$ . Each vertex moves 3 right and 5 down, so the image has vertices $(4, - 4)$ , $(7, - 4)$ and $(4, - 2)$ .
graph diagram — not drawn yet
A coordinate grid from -1 to 8 across and -5 to 4 up; a triangle with vertices A(1, 1), B(4, 1), C(1, 3) and its image with vertices A'(4, -4), B'(7, -4), C'(4, -2), both drawn with ruled sides; an arrow from A to A' labelled with the column vector 3 over -5.
Every vertex moves 3 right and 5 down, so the image is the same size, faces the same way and is not turned.
To describe a translation, choose one vertex and its image. Subtract: image minus object, $x$ first, then $y$ . Then check that the same vector works for a second vertex.
The trap here is subtracting the wrong way round. Object minus image gives the vector of the reverse move, with both signs wrong.
The reverse of a translation moves every point back to where it started. Change the sign of both numbers: the reverse of $(\binom{3}{- 5})$ is $(\binom{- 3}{5})$ .

Part 3 of 8

Reflection in a straight line

A reflection needs a mirror line. Each point and its image are the same distance from the line, on opposite sides of it. The line joining a point to its image crosses the mirror line at a right angle.
A point on the mirror line does not move. Its image is the point itself.
You need the equations of these lines. $x = a$ is a vertical line through $a$ on the $x$ -axis. $y = b$ is a horizontal line through $b$ on the $y$ -axis. The $x$ -axis is $y = 0$ and the $y$ -axis is $x = 0$ . The line $y = x$ goes up diagonally through $(0, 0)$ , $(1, 1)$ and $(2, 2)$ . The line $y = - x$ goes down diagonally through $(0, 0)$ , $(1, - 1)$ and $(2, - 2)$ .
The trap here is that $x = 3$ sounds as if it runs along the $x$ -axis. It does not. Every point on it has $x$ -coordinate 3, such as $(3, 0)$ and $(3, 5)$ , and those points are one above the other. So $x = 3$ is vertical.
For a vertical or horizontal mirror line, count squares at right angles to the line. To reflect $(1, 2)$ in $x = 3$ : the point is 2 squares to the left of the line, so its image is 2 squares to the right of it, at $(5, 2)$ . The $y$ -coordinate does not change.
For a diagonal mirror line, the right-angle crossing is diagonal too. Counting straight across or straight up gives the wrong point. Count diagonal steps to the line and the same number of diagonal steps beyond it, or use the coordinate rules in the next step.
Reflection in $y = x$ swaps the coordinates. Reflection in $y = - x$ swaps them and changes both signs. So $(4, 1)$ reflects in $y = x$ to $(1, 4)$ , and in $y = - x$ to $(- 1, - 4)$ .
$y = x : (x, y) \to (y, x) y = - x : (x, y) \to (- y, - x)$
graph diagram — not drawn yet
A coordinate grid from -1 to 5 both ways with the line y = x drawn and labelled; the point P(4, 1) and its image P'(1, 4) marked; a dashed segment from P to P' crossing y = x at (2.5, 2.5), with a right-angle mark at that crossing.
The join from P to its image meets the mirror line at a right angle, at the midpoint (2.5, 2.5). Counting straight across or straight up from P would give the wrong point.
To describe a reflection, find the mirror line. Join a point to its image and find the midpoint. Do the same for a second pair. Both midpoints lie on the mirror line, so the line through them is the mirror line. Give its equation.
The reverse of a reflection is the same reflection. Reflect the image in the same line and it lands back on the object.

Part 4 of 8

Drawing a rotation through $90 °$ , $180 °$ or $270 °$

A rotation needs three facts: the centre (the one point that does not move), the angle, and the direction, clockwise or anticlockwise. In this subtopic the angle is a multiple of $90 °$ : $90 °$ , $180 °$ or $270 °$ .
A $180 °$ turn ends in the same place whichever way you turn, so it needs no direction. A $270 °$ turn one way ends where a $90 °$ turn the other way ends: $90 °$ clockwise is the same as $270 °$ anticlockwise.
A reliable method is to turn the step from the centre to each vertex. Write the step as 'so many right, so many up'. Then turn the step.
Turned $90 °$ clockwise, a step of $a$ right and $b$ up becomes $b$ right and $a$ down. Turned $90 °$ anticlockwise, it becomes $b$ left and $a$ up. Turned $180 °$ , it becomes $a$ left and $b$ down.
Example: rotate the point $(4, 2)$ through $90 °$ clockwise about the centre $(1, 1)$ . The step from the centre is 3 right and 1 up. Turned clockwise it is 1 right and 3 down. Take that step from the centre: $(1 + 1, 1 - 3) = (2, - 2)$ .
graph diagram — not drawn yet
A coordinate grid from -1 to 5 across and -3 to 3 up; the centre C(1, 1) marked with a dot; the point P(4, 2) with a dashed step from C going 3 right then 1 up; the image P'(2, -2) with a dashed step from C going 1 right then 3 down; a curved arrow about C from P to P' showing a clockwise quarter turn.
The step from the centre turns with the point: 3 right and 1 up becomes 1 right and 3 down.
Do this for every vertex, then join the image vertices with a ruler. Check that each side of the image has the same length as the side it came from.
When the centre is $(0, 0)$ , the step is just the coordinates. That gives three rules you can use as a check.
$90 ° clockwise : (x, y) \to (y, - x) 90 ° anticlockwise : (x, y) \to (- y, x) 180 ° : (x, y) \to (- x, - y)$
The trap here is turning the shape on the spot, or about one of its own corners, when the centre is somewhere else. Only the centre stays still. Measure every step from it.
When you practise, tracing paper is a good check. Trace the object, hold the centre down with a pencil point, turn the paper a quarter turn and see where the shape lands.

Part 5 of 8

Describing a rotation, and its reverse

Find the angle and direction first. Pick a side of the object and look at the matching side of the image. A side that pointed right and now points down has turned $90 °$ clockwise. If it now points up, it has turned $90 °$ anticlockwise. If it now points left, it has turned $180 °$ .
Then find the centre. For a $180 °$ rotation, the centre is the midpoint of any point and its image. For example, if $(1, 5)$ goes to $(5, - 1)$ , the centre is $(3, 2)$ .
For a $90 °$ rotation, the centre is the same distance from a point as from its image. So it lies on the perpendicular bisector of the line joining them. Draw this for two pairs of points; the centre is where the two bisectors cross.
On a grid it can be quicker to try a likely point and test it. Take the step from your trial centre to one vertex, turn it, and see whether it lands on the image of that vertex. Test a second vertex too. If both land, you have the centre.
Write the description with all three facts: 'Rotation, $90 °$ clockwise, centre $(1, 1)$ '. For $180 °$ , 'Rotation, $180 °$ , centre $(3, 2)$ ' is complete.
The reverse of a rotation has the same centre and the same angle, in the opposite direction. The reverse of $90 °$ clockwise about $(1, 1)$ is $90 °$ anticlockwise about $(1, 1)$ . A $180 °$ rotation is its own reverse.

Part 6 of 8

Enlargement: positive and fractional scale factors

An enlargement needs two facts: the centre of enlargement and the scale factor. Every length of the image is the scale factor times the matching length of the object.
The step method works here too. Find the step from the centre to a vertex. Multiply both parts of the step by the scale factor $k$ . Take the new step from the centre. That gives the image vertex.
$image = centre + k \times (step from centre to point)$
Example: enlarge the point $(3, 3)$ with scale factor 2 and centre $(1, 2)$ . The step from the centre is 2 right and 1 up. Doubled, it is 4 right and 2 up. From the centre that gives $(1 + 4, 2 + 2) = (5, 4)$ .
The trap here is multiplying the coordinates by the scale factor. That measures from $(0, 0)$ , so it only works when the centre is $(0, 0)$ . With centre $(1, 2)$ , doubling $(3, 3)$ to $(6, 6)$ is wrong; the image is $(5, 4)$ .
Angles do not change in an enlargement, so the image is the same shape. A side 3 squares long becomes 6 squares long with scale factor 2.
A scale factor between 0 and 1 makes the image smaller and puts it nearer the centre. It is still called an enlargement. With scale factor $\frac{1}{2}$ and centre $(0, 0)$ , the point $(6, 4)$ goes to $(3, 2)$ , halfway to the centre.
To describe an enlargement, find the centre first. With a ruler, draw a straight line through each vertex and its image, and extend the lines. They all meet at one point. That point is the centre.
Then find the scale factor. Divide a length on the image by the matching length on the object.
$k = \frac{image length}{object length}$
Example: the triangle $(1, 1)$ , $(2, 1)$ , $(1, 3)$ has image $(3, 1)$ , $(6, 1)$ , $(3, 7)$ . The side from $(1, 1)$ to $(1, 3)$ is 2 long. Its image, from $(3, 1)$ to $(3, 7)$ , is 6 long. So $k = 6 \div 2 = 3$ .
graph diagram — not drawn yet
A coordinate grid from -1 to 7 across and 0 to 8 up; an object triangle with vertices (1, 1), (2, 1), (1, 3); an image triangle with vertices (3, 1), (6, 1), (3, 7); straight ruled lines from each image vertex back through the matching object vertex, all meeting at the point (0, 1), which is marked and labelled 'centre'.
Lines through matching vertices meet at the centre of enlargement, (0, 1). Each side of the image is 3 times the matching side of the object.
The line through $(1, 1)$ and $(3, 1)$ is $y = 1$ . The line through $(1, 3)$ and $(3, 7)$ meets $y = 1$ at $(0, 1)$ . So the description is: 'Enlargement, scale factor 3, centre $(0, 1)$ '.
Check it with the step method. From $(0, 1)$ to $(2, 1)$ is 2 right. Three times that is 6 right, which lands on $(6, 1)$ . That is the image of $(2, 1)$ , so the description is right.

Part 7 of 8

Negative scale factors

A negative scale factor multiplies the step by a negative number. That reverses the direction of the step. So the image is on the opposite side of the centre from the object.
Example: enlarge the triangle with vertices $(1, 1)$ , $(2, 1)$ and $(1, 2)$ with scale factor $- 2$ and centre $(0, 0)$ . The step to $(1, 2)$ is 1 right and 2 up. Multiplied by $- 2$ , it is 2 left and 4 down, so $(1, 2)$ goes to $(- 2, - 4)$ . In the same way $(1, 1)$ goes to $(- 2, - 2)$ and $(2, 1)$ goes to $(- 4, - 2)$ .
graph diagram — not drawn yet
A coordinate grid from -5 to 3 across and -5 to 3 up; the centre O(0, 0) marked; an object triangle with vertices (1, 1), (2, 1), (1, 2); an image triangle with vertices (-2, -2), (-4, -2), (-2, -4); ruled lines joining each object vertex to its image, all passing through O.
With a negative scale factor the lines from object to image cross at the centre, between the two shapes, and the image is upside down.
The lengths are multiplied by 2, the size of the scale factor without its sign. The minus sign also turns the image upside down, as if it had been rotated $180 °$ about the centre.
The centre lies between the object and the image. That is how you recognise a negative scale factor: the lines joining each vertex to its image cross between the two shapes.
A scale factor of $- 1$ gives an image of the same size, turned $180 °$ about the centre. That is the same as a rotation of $180 °$ about the same centre.
To describe a negative enlargement, find the centre where the lines cross. Divide an image length by the matching object length, and put a minus sign in front. An upside-down image half the size, on the far side of the centre, has scale factor $- \frac{1}{2}$ .
The reverse of an enlargement has the same centre and the reciprocal scale factor, $\frac{1}{k}$ . The reverse of scale factor 3 is $\frac{1}{3}$ . The reverse of $- \frac{1}{2}$ is $- 2$ . The sign does not change, because the image must come back through the centre to the object.

Part 8 of 8

Combinations of transformations

'Reflect in the $x$ -axis, then rotate…' means: do the first transformation, then do the second one to the image of the first.
Order matters. Take the point $(1, 2)$ . Reflect it in the $y$ -axis to get $(- 1, 2)$ , then translate by $(\binom{3}{0})$ to get $(2, 2)$ . The other way round, the translation gives $(4, 2)$ and the reflection then gives $(- 4, 2)$ . The two orders give different points.
To replace two transformations with a single one, follow the vertices through both. Then compare only the first shape with the last. Ignore the shape in the middle, and describe the single transformation in the usual way.
Example: reflect in the $x$ -axis, then in the $y$ -axis. The point $(x, y)$ goes to $(x, - y)$ , then to $(- x, - y)$ . That is the rule for a rotation of $180 °$ about $(0, 0)$ .
$(x, y) \to (x, - y) \to (- x, - y)$
Two reflections in parallel lines give a translation. Reflect $(2, 0)$ in $y = 1$ to get $(2, 2)$ , then reflect that in $y = 3$ to get $(2, 4)$ . The single transformation is a translation by $(\binom{0}{4})$ : twice the gap between the lines, in the direction from the first line to the second.
Two translations, one after the other, make one translation. Add the vectors: $(\binom{2}{1})$ then $(\binom{- 5}{3})$ is the same as $(\binom{- 3}{4})$ .
Here are the reverses together. Translation: change the sign of both numbers in the vector. Reflection: the same reflection. Rotation: the same centre and angle, the opposite direction. Enlargement: the same centre, scale factor $\frac{1}{k}$ .

Worked examples

Example 1. Triangle $P Q R$ has vertices $P (- 3, 4)$ , $Q (- 1, 4)$ and $R (- 3, 1)$ . It is translated by the vector $(\binom{5}{- 6})$ . (a) Find the coordinates of $P^{'}$ , $Q^{'}$ and $R^{'}$ . (b) Write down the column vector of the translation that maps $P^{'} Q^{'} R^{'}$ back onto $P Q R$ .

Show the working

Read the vector. The top number, 5, means 5 right. The bottom number, $- 6$ , means 6 down.
Add 5 to each $x$ -coordinate and add $- 6$ to each $y$ -coordinate.
$P^{'} = (- 3 + 5, 4 - 6) = (2, - 2)$
Do the same for the other two vertices.
$Q^{'} = (- 1 + 5, 4 - 6) = (4, - 2) R^{'} = (- 3 + 5, 1 - 6) = (2, - 5)$
Check a length. $P Q$ is horizontal and 2 long. $P^{'} Q^{'}$ , from $(2, - 2)$ to $(4, - 2)$ , is also horizontal and 2 long. A translation never changes a length or turns a side.
For the reverse, every point must move 5 left and 6 up. Change the sign of both numbers: $(\binom{- 5}{6})$ .

Answer: (a) $P^{'} (2, - 2)$ , $Q^{'} (4, - 2)$ , $R^{'} (2, - 5)$ . (b) $(\binom{- 5}{6})$ .

Example 2. Triangle $A$ has vertices $(2, 0)$ , $(4, 0)$ and $(4, 1)$ . Triangle $B$ has vertices $(0, 2)$ , $(0, 4)$ and $(1, 4)$ , in the matching order. Describe fully the single transformation that maps triangle $A$ onto triangle $B$ .

Show the working

Compare the sizes. The side from $(2, 0)$ to $(4, 0)$ is 2 long. The matching side, from $(0, 2)$ to $(0, 4)$ , is also 2 long. Same size, so it is not an enlargement.
It looks like a quarter turn about $(0, 0)$ , so test that. $90 °$ anticlockwise about $(0, 0)$ sends $(x, y)$ to $(- y, x)$ . That sends $(2, 0)$ to $(0, 2)$ and $(4, 0)$ to $(0, 4)$ . But it sends $(4, 1)$ to $(- 1, 4)$ , not $(1, 4)$ . So it is not that rotation. Two vertices that fit are not enough: test every vertex.
Test the direction of going round. In $A$ , the path $(2, 0) \to (4, 0) \to (4, 1)$ goes right, then up: anticlockwise. In $B$ , the path $(0, 2) \to (0, 4) \to (1, 4)$ goes up, then right: clockwise. The direction has reversed, so it is a reflection.
Find the mirror line from midpoints. The midpoint of $(2, 0)$ and $(0, 2)$ is $(1, 1)$ . The midpoint of $(4, 1)$ and $(1, 4)$ is $(2.5, 2.5)$ . Both points lie on the line $y = x$ .
Check with the rule for $y = x$ . Swap the coordinates of each vertex of $A$ : you get $(0, 2)$ , $(0, 4)$ and $(1, 4)$ . That is exactly triangle $B$ .

Answer: Reflection in the line $y = x$ .

Example 3. Triangle $P Q R$ has vertices $P (3, 2)$ , $Q (5, 2)$ and $R (3, 3)$ . (a) Find the vertices of its image after a rotation of $90 °$ clockwise about the centre $C (1, 1)$ . (b) Describe fully the single transformation that maps the image back onto triangle $P Q R$ .

Show the working

Write the step from the centre $C (1, 1)$ to each vertex. To $P$ : 2 right, 1 up. To $Q$ : 4 right, 1 up. To $R$ : 2 right, 2 up.
Turn each step $90 °$ clockwise: $a$ right and $b$ up becomes $b$ right and $a$ down. For $P$ : 1 right, 2 down. For $Q$ : 1 right, 4 down. For $R$ : 2 right, 2 down.
Take each new step from $C (1, 1)$ , not from the origin.
$P^{'} = (1 + 1, 1 - 2) = (2, - 1) Q^{'} = (1 + 1, 1 - 4) = (2, - 3) R^{'} = (1 + 2, 1 - 2) = (3, - 1)$
Check. $P Q$ goes 2 to the right. $P^{'} Q^{'}$ , from $(2, - 1)$ to $(2, - 3)$ , goes 2 down. A side that pointed right now points down, which is a clockwise quarter turn.
For the reverse, keep the centre and the angle, and change the direction.

Answer: (a) $P^{'} (2, - 1)$ , $Q^{'} (2, - 3)$ , $R^{'} (3, - 1)$ . (b) Rotation, $90 °$ anticlockwise (or $270 °$ clockwise), centre $(1, 1)$ .

Example 4. Triangle $T$ has vertices $(6, 1)$ , $(10, 1)$ and $(6, 9)$ . Find the vertices of the image of $T$ after an enlargement with scale factor $- \frac{1}{2}$ and centre $(2, 1)$ .

Show the working

Write the step from the centre $(2, 1)$ to each vertex. To $(6, 1)$ : 4 right, 0 up. To $(10, 1)$ : 8 right, 0 up. To $(6, 9)$ : 4 right, 8 up.
Multiply each step by $- \frac{1}{2}$ . The $\frac{1}{2}$ halves it and the minus sign reverses its direction. The new steps are: 2 left; 4 left; 2 left and 4 down.
Take each new step from the centre $(2, 1)$ .
$(2 - 2, 1) = (0, 1) (2 - 4, 1) = (- 2, 1) (2 - 2, 1 - 4) = (0, - 3)$
Check the size. The object side from $(6, 1)$ to $(6, 9)$ is 8 long. The image side from $(0, 1)$ to $(0, - 3)$ is 4 long, which is half of 8.
Check the position. The object is to the right of the centre and the image is to the left of it, upside down. That is what a negative scale factor does.

Answer: $(0, 1)$ , $(- 2, 1)$ and $(0, - 3)$ .

Where people go wrong

Mistake: Reading the column vector $(\binom{2}{- 3})$ as 2 up and 3 to the left.

Instead: The top number is the move across and the bottom number is the move up or down: $(\binom{2}{- 3})$ is 2 right and 3 down.

It follows the order of coordinates $(x, y)$ : the first number is always the $x$ direction.

Mistake: Reflecting in $y = x$ by counting squares straight across, or straight up, to the line.

Instead: Swap the coordinates, or count diagonal steps that cross the line at a right angle.

The join from a point to its image must meet the mirror line at a right angle. For a diagonal mirror line, that join is diagonal.

Mistake: Rotating the shape about one of its own corners when the centre is a different point.

Instead: Measure the step from the given centre to each vertex, and turn each step.

Only the centre stays still. Turning about the wrong point gives an image of the right shape, facing the right way, but in the wrong place.

Mistake: Multiplying the coordinates by the scale factor when the centre of enlargement is not $(0, 0)$ .

Instead: Multiply the step from the centre, then take the new step from the centre.

Multiplying the coordinates measures from $(0, 0)$ . With centre $(1, 2)$ and scale factor 2, the point $(3, 3)$ goes to $(5, 4)$ , not $(6, 6)$ .

Mistake: Leaving out a detail when you describe, for example writing 'rotation of $90 °$ ' with no centre or direction.

Instead: Give every detail the type needs. A rotation needs the centre, the angle and the direction.

A $90 °$ rotation about a different centre puts the image somewhere else, so without the centre the description does not fix one transformation.

Check yourself

The point $(3, - 2)$ is translated by $(\binom{- 4}{5})$ . Find its image.
Answer
$(3 - 4, - 2 + 5) = (- 1, 3)$ .
A translation maps the point $(1, 4)$ onto the point $(6, 1)$ . Write down its column vector, and the column vector of the reverse translation.
Answer
Image minus object: $(\binom{6 - 1}{1 - 4}) = (\binom{5}{- 3})$ . The reverse is $(\binom{- 5}{3})$ .
Reflect the point $(5, 2)$ in the line $y = - 1$ .
Answer
The point is 3 squares above the line, so its image is 3 squares below it: $(5, - 4)$ .
Reflect the point $(3, - 1)$ in the line $y = x$ , and then (separately) in the line $y = - x$ .
Answer
In $y = x$ , swap the coordinates: $(- 1, 3)$ . In $y = - x$ , swap them and change both signs: $(1, - 3)$ .
Rotate the point $(4, 1)$ through $90 °$ anticlockwise about the centre $(2, 2)$ .
Answer
The step from the centre is 2 right and 1 down. Turned $90 °$ anticlockwise it is 1 right and 2 up. From $(2, 2)$ that gives $(3, 4)$ .
A rotation of $180 °$ maps the point $(1, 5)$ onto the point $(5, - 1)$ . Find the centre of the rotation.
Answer
The midpoint of the point and its image: $(\frac{1 + 5}{2}, \frac{5 + (- 1)}{2}) = (3, 2)$ .
Enlarge the point $(3, 1)$ with scale factor 3 and centre $(1, 2)$ . Then write down the scale factor of the reverse enlargement.
Answer
The step is 2 right and 1 down; three times it is 6 right and 3 down. From $(1, 2)$ that gives $(7, - 1)$ . The reverse has scale factor $\frac{1}{3}$ , centre $(1, 2)$ .
A shape is reflected in the $x$ -axis, and the image is then reflected in the $y$ -axis. Describe fully the single transformation that has the same effect.
Answer
$(x, y) \to (x, - y) \to (- x, - y)$ , so it is a rotation, $180 °$ , centre $(0, 0)$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Use a coordinate grid with $x$ from $- 6$ to $6$ and $y$ from $- 6$ to $6$ . Triangle $A$ has vertices $(2, 1)$ , $(5, 1)$ and $(2, 3)$ .

(a) Draw the image of triangle $A$ after a reflection in the line $y = - 1$ . Label it $B$ . 2 marks No calculator

Mark scheme

B2 triangle with vertices $(2, - 3)$ , $(5, - 3)$ and $(2, - 5)$ , sides ruled; or B1 for a reflection of $A$ in a horizontal line other than $y = - 1$ , or for two vertices correct

(b) Draw the image of triangle $A$ after a rotation of $90 °$ clockwise about the point $(0, 2)$ . Label it $C$ . 2 marks No calculator

Mark scheme

B2 triangle with vertices $(- 1, 0)$ , $(- 1, - 3)$ and $(1, 0)$ , sides ruled; or B1 for a $90 °$ clockwise rotation of $A$ about a different centre, or a $90 °$ anticlockwise rotation about $(0, 2)$

(c) Triangle $D$ has vertices $(- 1, 2)$ , $(- 1, 5)$ and $(- 3, 2)$ . Describe fully the single transformation that maps triangle $A$ onto triangle $D$ . 3 marks No calculator

Mark scheme

B1 Rotation (an answer naming more than one transformation scores no marks)
B1 $90 °$ anticlockwise oe, e.g. $270 °$ clockwise
B1 centre $(0, 0)$ , or centre the origin

Question 2. Use a coordinate grid with $x$ from $- 6$ to $7$ and $y$ from $- 6$ to $7$ . Triangle $P$ has vertices $(2, 2)$ , $(6, 2)$ and $(6, 4)$ .

(a) Draw the image of triangle $P$ after an enlargement with scale factor $\frac{1}{2}$ and centre $(- 2, - 2)$ . 2 marks No calculator

Mark scheme

B2 triangle with vertices $(0, 0)$ , $(2, 0)$ and $(2, 1)$ , sides ruled; or B1 for a triangle of the correct size and orientation (sides 2 across and 1 up) in the wrong position, or for two vertices correct

(b) Triangle $Q$ has vertices $(- 1, - 1)$ , $(- 3, - 1)$ and $(- 3, - 2)$ . Describe fully the single transformation that maps triangle $P$ onto triangle $Q$ . 3 marks No calculator

Mark scheme

B1 Enlargement
B1 scale factor $- \frac{1}{2}$ oe, e.g. $- 0.5$
B1 centre $(0, 0)$

(c) Describe fully the single transformation that maps triangle $Q$ onto triangle $P$ . 2 marks No calculator

Mark scheme

B1 Enlargement, centre $(0, 0)$
B1 scale factor $- 2$ ; FT the reciprocal of their scale factor in (b)

Question 3. Triangle $R$ has vertices $(1, 2)$ , $(3, 2)$ and $(3, 5)$ .

(a) Triangle $R$ is translated by the vector $(\binom{- 4}{1})$ to give triangle $S$ . Write down the coordinates of the vertices of triangle $S$ . 2 marks No calculator

Mark scheme

B2 $(- 3, 3)$ , $(- 1, 3)$ and $(- 1, 6)$ ; or B1 for two vertices correct, or for all three vertices translated by $(\binom{1}{- 4})$

(b) Triangle $S$ is then translated by the vector $(\binom{6}{- 3})$ to give triangle $U$ . Write down the column vector of the single translation that maps triangle $U$ onto triangle $R$ . 1 mark No calculator

Mark scheme

B1 $(\binom{- 2}{2})$ cao

(c) Triangle $R$ is reflected in the line $x = 4$ . The image is then reflected in the line $x = 7$ . Describe fully the single transformation that maps triangle $R$ onto the final image. 2 marks No calculator

Mark scheme

B1 Translation
B1 $(\binom{6}{0})$

How this is examined

Describing fully the single transformation that maps one shape onto another: A Describe fully worth about 3 marks, with the object and the image both drawn on a grid and nothing else given The name of one transformation, then every detail that type needs: 'Rotation, $90 °$ anticlockwise, centre $(0, 0)$ '; 'Reflection in the line $y = - x$ '; 'Translation by $(\binom{- 4}{1})$ '. The marks are for the name and for each detail, so a rotation with no centre, or a reflection with no line, does not score in full. Two transformations named together do not answer a question that asks for a single one.
Finding the centre and the scale factor of an enlargement, including a fractional or a negative scale factor: A 3-mark Describe fully where the image is smaller than the object, or on the other side of the centre The word 'Enlargement'; the centre as coordinates, found where ruled lines through matching vertices meet; and the scale factor as image length ÷ object length, with a minus sign when the centre lies between the two shapes. For example: 'Enlargement, scale factor $- \frac{1}{2}$ , centre $(0, 0)$ '.
Drawing the image of a shape accurately on a grid: A Draw of about 2 marks: a reflection in a line such as $y = - 1$ or $y = x$ , a rotation about a point away from the shape, or an enlargement with a given centre Every vertex at the correct grid point, with the sides ruled. Work out each vertex from the mirror line or the centre with the step method, then check the side lengths and which way the image faces. A shape of the right size in the wrong place is not the image asked for.
Replacing two transformations, done one after the other, with a single transformation: A Describe fully of 2 or 3 marks, after you have drawn or worked out the two images in turn The first shape compared with the last shape only, then a full description of that one transformation. For example, a reflection in the $x$ -axis followed by a reflection in the $y$ -axis is 'Rotation, $180 °$ , centre $(0, 0)$ '.
Giving the reverse of a transformation: A Write down or Describe of 1 or 2 marks, after a part where you carried out the transformation The same type with the same line or centre, and the change written out: the vector with both signs changed, the opposite direction of turn, or the reciprocal scale factor ( $3$ becomes $\frac{1}{3}$ , $- \frac{1}{2}$ becomes $- 2$ ). A reflection is its own reverse.

Words to know

Object: The shape before a transformation.
Image: The shape after a transformation. Its vertices are labelled $A^{'}$ , $B^{'}$ , $C^{'}$ to match the object's $A$ , $B$ , $C$ .
Translation: A slide in which every point moves by the same column vector.
Column vector: $(\binom{x}{y})$ : the top number is the move right (positive) or left (negative); the bottom number is the move up (positive) or down (negative).
Reflection: A flip in a mirror line. Each point and its image are the same distance from the line, and the join between them crosses the line at a right angle.
Rotation: A turn about a fixed point, the centre of rotation, through a given angle in a given direction.
Enlargement: A change of size from a fixed point, the centre of enlargement, by a scale factor. A scale factor between 0 and 1 makes the image smaller; a negative scale factor puts the image on the other side of the centre, upside down.
Scale factor: The number that every step from the centre is multiplied by in an enlargement. Its size, without the sign, is image length ÷ object length.
Congruent: The same shape and the same size. Translations, reflections and rotations give congruent images.

All Maths lessons Next: 8.2 Vectors in two dimensions

8.1 Transformations

Object, image, and what each transformation needs

Translation by a column vector

Reflection in a straight line

Drawing a rotation through 90°, 180° or 270°

Describing a rotation, and its reverse

Enlargement: positive and fractional scale factors

Negative scale factors

Combinations of transformations

Worked examples

Where people go wrong

Check yourself

Exam questions

How this is examined

Words to know

Drawing a rotation through $90 °$ , $180 °$ or $270 °$