Composite rotations and mixed transformations – Inverse and Composite Transformations in 2D (Senior 3 Mathematics) | Ganzaa

Inverse and Composite Transformations in 2D

Lessons

Senior 3 · Mathematics

Inverse and Composite Transformations in 2D

4 lessons

Lesson 3 of 4

Composite rotations and mixed transformations

25 min

Turning, then turning again

From 12:00 to 12:15 the minute hand of a clock turns through $90^\circ$ ; from 12:15 to 12:35 it turns a further $120^\circ$ . Altogether it has turned $210^\circ$ , because both turns are about the same centre. Turns about different centres, or a turn followed by a flip, are less obvious. This lesson finds the single transformation in each case, and shows that the order in which you do them can change the answer.

Definition

A glide reflection is a reflection in a line $m$ followed by a translation by a vector parallel to $m$ . Footprints in sand, left, right, left, right, are images of one another under glide reflections.

Example: $(x, y) \mapsto (x + 4,\; -y)$ is a glide reflection: reflect in the $x$ -axis, then slide 4 units along it.

Important

1. Rotations about the same centre add: $R_{X, \beta} \circ R_{X, \alpha} = R_{X, \alpha + \beta}$ . In particular $S_X \circ S_X$ is the identity.

2. Two half-turns about different centres $A$ then $B$ give a translation by $2\overrightarrow{AB}$ : $S_B \circ S_A = T_{2\overrightarrow{AB}}$ .

3. Rotations about different centres: the angles still add. The result is a rotation through $\alpha + \beta$ about a new centre, or a translation if $\alpha + \beta$ is a whole number of turns.

4. Mixed pairs: a reflection and a translation give a glide reflection (translation along the mirror) or a reflection in a parallel line (translation perpendicular to the mirror); a rotation and a reflection give a reflection if the centre of rotation is on the mirror, and a glide reflection if it is not. In general $g \circ f \neq f \circ g$ : the order matters.

Worked example 1: two rotations about the origin

Find the single transformation equal to $R_{O, 90^\circ}$ followed by $R_{O, 180^\circ}$ , and the image of $(3, 1)$ .

Solution The angles add: $R_{O, 270^\circ} = R_{O, -90^\circ}$ , with rule $(x, y) \mapsto (y, -x)$ . Step by step: $(3, 1) \mapsto (-1, 3) \mapsto (1, -3)$ , and $(y, - x) = (1, - 3)$ . ✓

Worked example 2: two half-turns

Triangle $P (0, 0)$ , $Q (2, 0)$ , $R (0, 1)$ is given a half-turn about $A (1, 1)$ and then a half-turn about $B (4, 2)$ . Find the images and the single transformation.

Solution $S_A : (x, y) \mapsto (2 - x,\; 2 - y)$ gives $P^{'} (2, 2)$ , $Q^{'} (0, 2)$ , $R^{'} (2, 1)$ . $S_B : (x, y) \mapsto (8 - x,\; 4 - y)$ gives $P^{''} (6, 2)$ , $Q^{''} (8, 2)$ , $R^{''} (6, 3)$ .

In general $(x, y) \mapsto (2 - x,\; 2 - y) \mapsto \big(8 - (2 - x),\; 4 - (2 - y)\big) = (x + 6,\; y + 2)$ : the translation by $\begin{pmatrix} 6 \\ 2 \end{pmatrix} = 2\overrightarrow{AB}$ , since $\overrightarrow{AB} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ .

Worked example 3: quarter-turns about two different centres

Describe $R_{Y, 90^\circ} \circ R_{O, 90^\circ}$ , where $Y (2, 0)$ .

Solution $R_{O, 90^\circ} : (x, y) \mapsto (-y, x)$ . For $R_{Y, 90^\circ}$ , turn the vector from $Y$ : $\begin{pmatrix} -y - 2 \\ x \end{pmatrix} \mapsto \begin{pmatrix} -x \\ -y - 2 \end{pmatrix}$ , and add $Y$ : the point $(2 - x,\; -y - 2)$ .

So the composite is $(x, y) \mapsto (2 - x,\; -2 - y) = (2 \times 1 - x,\; 2 \times (-1) - y)$ : the half-turn $S_{(1, -1)}$ , a rotation through $90^\circ + 90^\circ = 180^\circ$ about the new centre $(1, - 1)$ .

Worked example 4: the order matters

Apply $R_{O, 90^\circ}$ and the reflection in the $x$ -axis to triangle $A (1, 2)$ , $B (3, 2)$ , $C (1, 3)$ , in both orders.

Solution Rotate first, then reflect: $(x, y) \mapsto (-y, x) \mapsto (-y, -x)$ , the reflection in $y = - x$ . Images: $(- 2, - 1)$ , $(- 2, - 3)$ , $(- 3, - 1)$ (orange).

Reflect first, then rotate: $(x, y) \mapsto (x, -y) \mapsto (y, x)$ , the reflection in $y = x$ . Images: $(2, 1)$ , $(2, 3)$ , $(3, 1)$ (green). Different orders, different answers.

Worked example 5: a reflection and a translation

Describe (a) the reflection in the $x$ -axis followed by $T_{(4, 0)}$ ; (b) the reflection in the $x$ -axis followed by $T_{(0, 6)}$ .

Solution (a) $(x, y) \mapsto (x, -y) \mapsto (x + 4,\; -y)$ : a glide reflection along the $x$ -axis, 4 units to the right (shown above).

(b) $(x, y) \mapsto (x, -y) \mapsto (x,\; 6 - y) = (x,\; 2 \times 3 - y)$ : the reflection in the line $y = 3$ . A translation at right angles to the mirror just moves the mirror, by half the vector.

Warning

Read $g \circ f$ from right to left: $f$ is done first. Doing the steps in the written order, left to right, is the most common error, and with a rotation and a reflection it gives a different mirror line.

Tip

Count the reflections. An even number of reflections (including none) gives a translation or a rotation; an odd number gives a reflection or a glide reflection. This tells you which answers are possible before you work anything out.

SaenzMath: Glide reflection, transformation composition. Watch how a reflection and a translation along the mirror combine into one move, as in Worked example 5 (a).

✦

Practice

The last two are harder.

Find the single transformation equal to $R_{O, 60^\circ}$ followed by $R_{O, -150^\circ}$ .
$A (0, 0)$ and $B (3, - 1)$ . Find the single transformation equal to $S_{A}$ followed by $S_{B}$ , and the image of $(1, 1)$ .
Describe the reflection in the $y$ -axis followed by $T_{(0, 5)}$ , and find the image of $(2, 1)$ .
Show that the reflection in the $y$ -axis followed by $T_{(6, 0)}$ is a reflection, and give its mirror line.
Describe the reflection in $y = x$ followed by $R_{O, 180^\circ}$ .
Show that $T_{(2, 0)}$ followed by $R_{O, 90^\circ}$ is a rotation through $90^\circ$ , and find its centre.
Explain why $S_B \circ S_A$ and $S_A \circ S_B$ are different when $A \neq B$ .

Show answersHide answers

$R_{O, -90^\circ}$ , a quarter-turn clockwise about $O$ .
$T_{2\overrightarrow{AB}}$ with $2\overrightarrow{AB} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}$ ; $(1, 1) \mapsto (7, -1)$ .
$(x, y) \mapsto (-x,\; y + 5)$ : a glide reflection along the $y$ -axis, 5 units up; $(2, 1) \mapsto (-2, 6)$ .
$(x, y) \mapsto (-x, y) \mapsto (6 - x,\; y) = (2 \times 3 - x,\; y)$ : the reflection in $x = 3$ .
$(x, y) \mapsto (y, x) \mapsto (-y, -x)$ : the reflection in $y = - x$ .
$(x, y) \mapsto (x + 2,\; y) \mapsto (-y,\; x + 2)$ . The centre does not move: $- y = x$ and $x + 2 = y$ give $(- 1, 1)$ . It is $R_{(-1, 1), 90^\circ}$ .
$S_B \circ S_A = T_{2\overrightarrow{AB}}$ but $S_A \circ S_B = T_{2\overrightarrow{BA}}$ ; these vectors are opposite, so the translations go in opposite directions.

Sources and credits

Written by Ganzaa.
Video: Glide Reflection - Transformation Composition, SaenzMath, embedded from YouTube.

The diagrams, worked examples and practice items were written by Ganzaa.

1Turning, then turning again

Turning, then turning again