Transformations

Mathematics revision notes · Cambridge (CIE) · 0580

1. The four transformations

Object, image and what changes

The starting shape is the object and the result is the image, with the image's vertices labelled A′, B′, C′ to match A, B, C
Three of the four keep the shape exactly the same size, so object and image are congruent: reflection, rotation and translation
Only enlargement changes the size, and even then the shape stays similar
A point that does not move under a transformation is invariant
A ruler must be used for every straight edge, which the syllabus states directly

What each transformation needs

Transformation	What a question can ask for
Reflection	a mirror line that is any straight line, including y = x and y = −x
Rotation	any centre, through a multiple of 90°
Enlargement	positive, fractional and negative scale factors
Translation	a column vector
Combinations	two transformations one after the other, described as a single one

2. Reflection

Doing one

A reflection flips the shape across a mirror line, so the image is the same size but the wrong way round
Every point and its image sit the same perpendicular distance from the mirror line, on opposite sides

A square object ABCD and its image across a dashed vertical mirror line, each vertex marked with a cross: A sits nearest the line and its image A prime sits the same number of squares on the other side, and the labelling order runs the opposite way round — Source: Reflections by Save My Exams

Take one vertex at a time, count squares straight across to the line at right angles, then continue the same count past it

A shape reflected in the dashed line x = −1, with dashed guide lines drawn from two of its vertices straight across to the mirror line and the same distance again beyond it, each meeting the line at a marked right angle — Source: Reflections by Save My Exams

For a diagonal mirror line such as y = x, count diagonally across the squares rather than horizontally
Any vertex sitting on the line itself stays exactly where it is, making it invariant
Where the mirror line passes through the shape, part of it reflects each way

Worked example

Reflecting in the line y = x

Triangle A has vertices (−1, 6), (3, 6) and (3, 8). Reflect A in the line y = x and draw its image.

Solution:

Draw the mirror line first, with a ruler: y = x runs diagonally through (0, 0), (1, 1), (2, 2) and so on
A reflection in y = x swaps the two coordinates of every point, so (a, b) goes to (b, a)
(−1, 6) goes to (6, −1), (3, 6) goes to (6, 3), and (3, 8) goes to (8, 3)
Check one pair: (−1, 6) and (6, −1) have midpoint (2.5, 2.5), which lies on y = x, so they sit the same distance either side of the line
Plot the three points and join them with a ruler: the image has vertices (6, −1), (6, 3) and (8, 3)
Swapping the coordinates and also changing both signs would reflect in y = −x instead, sending (−1, 6) to (−6, 1)

A grid with the dashed mirror line y = x. Triangle A has vertices (−1, 6), (3, 6) and (3, 8), and its reflection A′ has vertices (6, −1), (6, 3) and (8, 3), on the other side of the line; a dotted guide joins (−1, 6) to (6, −1), crossing the mirror at right angles at (2.5, 2.5)

Describing one

Say reflection, then state which line the shape was flipped across, written as an equation
Vertical lines are x = k and horizontal lines are y = k, which is the pair most often swapped over
The diagonal through the origin sloping up is y = x, and the one sloping down is y = −x

Worked example

Describing a reflection fully

Triangle A has vertices (1, 2), (3, 2) and (1, 5). Its image B has vertices (−1, 2), (−3, 2) and (−1, 5). Describe fully the single transformation that maps A onto B.

A coordinate grid with x from −4 to 4 and y from 0 to 6. Triangle A is drawn with vertices at (1, 2), (3, 2) and (1, 5), and triangle B with vertices at (−1, 2), (−3, 2) and (−1, 5). No mirror line is drawn.

Solution:

The shape is the same size and the same way up, but it has been flipped, so this is a reflection
Compare one pair of matching points: (1, 2) becomes (−1, 2)
The y coordinate is unchanged and the x coordinate has changed sign
The mirror line is therefore vertical, midway between 1 and −1, which is x = 0
Check on a second pair: (3, 2) becomes (−3, 2), midway again at x = 0
The answer is a reflection in the line x = 0
Give the mirror line as its equation: x = 0 is the y-axis

3. Rotation

Doing one

A rotation turns the shape about a fixed point, the centre of rotation, through a multiple of 90°
Tracing paper may be used in the examination (ask for it, since you cannot bring your own), and it is by far the most reliable method
- Trace the object, press a pencil point onto the centre, turn the paper through the given angle, then copy the image across
- Drawing an arrow on the tracing paper first shows at a glance how far it has turned

Before rotating: the centre (0, 2) is marked with a red cross beside a curved arrow showing an anticlockwise turn, and the vertex (3, 1) of triangle A is marked, with (1, 5), where a quarter turn anticlockwise takes it — Source: Rotations by Save My Exams

The size and shape are unchanged, only the position and the orientation
If the centre lies on the shape, that point is invariant

Shape A rotated clockwise about a centre of rotation marked at (−4, 0), red lines join the centre to the vertex (−4, 4) and to its image (0, 0) at right angles, and a curved arrow shows the turn to the image B on the other side of the y-axis — Source: Rotations by Save My Exams

Describing one

Say rotation, then give the angle with its direction, then the centre as coordinates

Two clock faces with a red curved arrow on each: the first turns the way the hands move and is labelled clockwise, the second turns against them and is labelled anticlockwise — Source: Rotations by Save My Exams

Half turns need no direction, since 180° clockwise and anticlockwise give the same image
Mark schemes accept equivalents such as 270° clockwise for 90° anticlockwise, and abbreviations like acw

Worked example

Describing a rotation

Triangle P has vertices (1, 1), (1, 3) and (2, 1). Triangle Q has vertices (−1, 1), (−3, 1) and (−1, 2). Describe fully the single transformation that maps P onto Q.

A coordinate grid with x from −4 to 3 and y from 0 to 4. Triangle P is drawn with vertices at (1, 1), (1, 3) and (2, 1), and triangle Q with vertices at (−1, 1), (−3, 1) and (−1, 2). No centre of rotation is marked.

Solution:

P and Q are the same size and shape, and Q is P turned, not flipped: from the right angle, P's longer side points up and its shorter side right, while Q's longer side points left and its shorter side up, so this is a rotation
Tracing P and turning it shows a quarter turn fits, so the angle is 90°
Turning anticlockwise takes the vertex (1, 3) round to (−3, 1), which matches Q
Testing the origin as the centre: it is the same distance from (1, 1) as from (−1, 1), and likewise for the other pairs
The full description is a rotation, 90° anticlockwise, centre (0, 0)
Check: a 90° anticlockwise turn about the origin sends (x, y) to (−y, x), and (1, 1), (1, 3), (2, 1) do map to (−1, 1), (−3, 1), (−1, 2)

4. Translation

Doing and describing one

A translation slides the shape without turning or resizing it, so the orientation is unchanged
The movement is given as a column vector, with the top number the horizontal move and the bottom number the vertical one
Positive top means right and negative means left; positive bottom means up and negative means down
Move every vertex by the same vector, then join them up

Shape P translated to its image P prime, with a red stepped path drawn from one vertex of P to the matching vertex of P prime, counting the horizontal steps first and then the vertical ones — Source: Translations by Save My Exams

To describe one, say translation and give the vector
Mark schemes also accept the movement in words, such as "2 right, 7 down"

Shape A translated to shape B, with a red stepped path drawn between one pair of matching vertices and annotated 2 right and 3 down, which is the column vector of the translation — Source: Translations by Save My Exams

Worked example

Translating by a column vector

Triangle A has vertices (5, 1), (9, 1) and (9, 3). Draw the image of A after a translation by the column vector with top number −4 and bottom number 5.

Solution:

The top number −4 means 4 squares to the left, and the bottom number 5 means 5 squares up
Move every vertex by the same amount: (5, 1) goes to (5 − 4, 1 + 5) = (1, 6)
In the same way, (9, 1) goes to (5, 6) and (9, 3) goes to (5, 8)
Join the three points with a ruler: the image has vertices (1, 6), (5, 6) and (5, 8)
Check: the image is the same size, shape and way round as A, and its base from (1, 6) to (5, 6) is horizontal and 4 long, like A's base from (5, 1) to (9, 1)

Triangle A with vertices (5, 1), (9, 1) and (9, 3) and its image A′ with vertices (1, 6), (5, 6) and (5, 8), with arrows from (5, 1) going 4 squares left and then 5 squares up to (1, 6)

5. Enlargement

Doing one

An enlargement resizes the shape by a scale factor from a fixed centre of enlargement
Each vertex moves along the line from the centre through it, ending scale-factor times as far from the centre
Multiply each distance from the centre, both across and up, by the scale factor

Shape C enlarged to C prime with the centre of enlargement marked at (2, 1): a red ray is drawn from the centre through a vertex of C and on to the matching vertex of C prime, which sits twice as far along it — Source: Enlargements by Save My Exams

A scale factor between 0 and 1 makes the image smaller, even though it is still called an enlargement

A large shape enlarged by a fraction to a much smaller green image: the rays from the centre of enlargement at (4, 2) run from each vertex of the object inwards, and the image sits between the object and the centre — Source: Enlargements by Save My Exams

A negative scale factor throws the image across to the far side of the centre and inverts it, so it acts as a 180° rotation as well as a resize

Triangle A enlarged with scale factor −2 about the centre marked with a cross at (1, 2): rays run from each vertex of A through the centre and out the other side, so the image B is twice as large, upside down and on the opposite side of the centre

Every length multiplies by the scale factor, so the image is similar to the object rather than congruent

Describing one

Say enlargement, then give the scale factor, then the centre as coordinates
Find the centre by drawing a line through each pair of corresponding vertices and seeing where they all cross

Working backwards to find a centre of enlargement: a red line is ruled through three pairs of matching vertices of the large shape A and the small shape B, and all of the lines meet at one point, which is the centre — Source: Enlargements by Save My Exams

Find the scale factor by dividing an image length by the matching object length

Worked example

Describing an enlargement

Triangle A has vertices (1, 1), (3, 1) and (1, 2). Triangle B has vertices (2, 2), (6, 2) and (2, 4). Describe fully the single transformation that maps A onto B.

A coordinate grid with x from 0 to 7 and y from 0 to 5. Triangle A is drawn with vertices at (1, 1), (3, 1) and (1, 2), and the larger triangle B with vertices at (2, 2), (6, 2) and (2, 4). No centre of enlargement and no rays are drawn.

Solution:

B is bigger than A, so the only candidate is an enlargement: the other three keep the size the same
Compare a pair of matching sides: A has a base from (1, 1) to (3, 1), which is 2 units, and B has a base from (2, 2) to (6, 2), which is 4 units
Scale factor = 4 ÷ 2 = 2
Draw a line through (1, 1) and (2, 2), another through (3, 1) and (6, 2), and a third through (1, 2) and (2, 4)
All three meet at the origin, so that is the centre
The full description is an enlargement, scale factor 2, centre (0, 0)
Check: each vertex of B is exactly twice as far from the origin as the matching vertex of A

Exam tip

One transformation, every property

Name exactly one transformation, in its proper word: rotation, not turn or spin; enlargement, not dilation or reduction. Naming two scores nothing for the whole part, so a negative scale factor is one enlargement, never a rotation plus an enlargement. The name and each property (angle with direction, centre, scale factor, vector) earn a mark each, so write every one even when unsure of one. Describing a single transformation is set in 11 of the 21 papers.

Worked example

An enlargement with a negative scale factor

Triangle A has vertices (−1, 2), (−3, 2) and (−1, 6). Triangle B has vertices (5, −1), (6, −1) and (5, −3). Describe fully the single transformation that maps A onto B.

A coordinate grid with x from −4 to 7 and y from −4 to 7. Triangle A is drawn with vertices at (−1, 2), (−3, 2) and (−1, 6), and the smaller triangle B with vertices at (5, −1), (6, −1) and (5, −3). No centre of enlargement and no rays are drawn.

Solution:

B is smaller than A, so this is an enlargement, the only one of the four transformations that changes size
Compare matching sides: A's vertical side from (−1, 2) to (−1, 6) is 4 units, and B's from (5, −1) to (5, −3) is 2 units, so the size factor is 2 ÷ 4 = ½
B is upside down: from the right angle, A's sides run up and to the left, while B's run down and to the right, so the image is inverted and the scale factor is negative, −½
Join each pair of matching vertices: (−1, 2) to (5, −1), (−3, 2) to (6, −1), and (−1, 6) to (5, −3)
The three lines cross at (3, 0), between the two shapes, which is where the centre of a negative enlargement always lies
Check from the centre: (−1, 2) is 4 left and 2 up from (3, 0), and −½ of that is 2 right and 1 down, which lands on (5, −1)
The full description is an enlargement, scale factor −½, centre (3, 0)