Transformation

A transformation changes the position of a point or a shape by a fixed rule. The shape you start with is the object; where it ends up is its image. If a point P goes to P′, we write P → P′ (read “P maps to P-dash”).

The four you know already:

• Reflection flips a shape over a mirror line. Each point and its image are the same distance from the line, on opposite sides.
• Rotation turns a shape about a fixed point, the center of rotation, through an angle. An anticlockwise turn is positive (+90∘°), a clockwise turn is negative (−90∘°).
• Translation slides a shape: every point moves the same distance in the same direction.
• Enlargement (also called dilation) makes a shape bigger or smaller from a fixed point, the center of enlargement, by a scale factor k.

Reflection, rotation and translation never change the size or shape: the image is congruent to the object. They are called isometric transformations (iso means “same”, metric “measure”). An enlargement keeps the shape but changes the size, so its image is similar to the object, not congruent.

The rules from Grade 9

This unit uses all of these, so keep the table handy. Each says where a point P(x, y) goes.

Transformation	Image of P(x, y)
reflection in the X-axis (y = 0)	P′(x, −y)
reflection in the Y-axis (x = 0)	P′(−x, y)
reflection in the line y = x	P′(y, x)
reflection in the line y = −x	P′(−y, −x)
reflection in the line x = h	P′(2h − x, y)
reflection in the line y = k	P′(x, 2k − y)
rotation +90∘° about O(0, 0)	P′(−y, x)
rotation 180∘° about O(0, 0)	P′(−x, −y)
rotation −90∘° (the same as +270∘°) about O(0, 0)	P′(y, −x)
rotation +90∘° about (a, b)	P′(a − (y − b), b + (x − a))
rotation 180∘° about (a, b)	P′(2a − x, 2b − y)
rotation −90∘° about (a, b)	P′(a + (y − b), b − (x − a))
translation by $\begin{pmatrix} a \\ b \end{pmatrix}$	P′(x + a, y + b)
enlargement E[(0, 0), k]	P′(kx, ky)
enlargement E[(a, b), k]	P′(k(x − a) + a, k(y − b) + b)

The column of two numbers is the translation vector. Its top number says how far to move along the X-axis, its bottom number how far along the Y-axis:

• right is +, left is −;
• up is +, down is −.

So a translation by the vector $T = \begin{pmatrix} a \\ b \end{pmatrix}$ adds a to every x-coordinate and b to every y-coordinate:

$$P(x, y) \xrightarrow{\ T = \begin{pmatrix} a \\ b \end{pmatrix}\ } P'(x + a,\ y + b)$$

Moving every point of a shape the same distance in the same direction is called translation. A translation has both a size, its magnitude (how far), and a direction (which way), so it is a vector. It is fully described by its translation vector.

Worked examples

Question 1

Question 2

Questions 3 to 13

In every case, two moves one after the other could be replaced by one move. When a figure is translated from A to A′ by one translation and then from A′ to A′′ by another, the single translation that takes it straight from A to A′′ is called the combined translation. The same idea works for every kind of transformation.

9.3.1 Combining two transformations of the same kind

When two transformations are done one after the other, the single transformation that has the same effect is their combined (or composite) transformation, also called the composition of the two. In this section both are of the same kind: two reflections, two rotations, two translations or two enlargements.

If F is done first and G second, the combined transformation is written G ∘ F (read “G after F”, or “G of F”):

$$(G \circ F)(P) = G\big(F(P)\big)$$

It works exactly like the composite function g ∘ f of Unit 1: the transformation written nearest the point acts first.

Two reflections in lines that cross

In Figure 9.21, P(5, −2) is reflected in the X-axis (y = 0) to P′(5, 2), and P′ is then reflected in the line y = x to P′′(2, 5).

The rotation rule for +90∘° about O is (x, y) → (−y, x), and indeed (5, −2) → (2, 5) = P′′. So the two reflections together are a rotation of +90∘° about O, and 90∘° is twice the 45∘° between the mirrors.

Why twice? A reflection keeps every point the same distance from any point on the mirror, so P, P′ and P′′ are all the same distance from O, where the mirrors cross: P has turned about O. Now follow the angles. P is some angle α below the first mirror; its reflection P′ is the same angle α above it. P′ is then some angle β below the second mirror, and P′′ the same β above it. Altogether P has turned α + α + β + β = 2(α + β), and α + β is exactly the angle from the first mirror to the second.

Two reflections in lines that cross make a rotation:

• its center is the point where the mirrors cross;
• its angle is twice the angle between the mirrors;
• its direction is from the first mirror towards the second.

Two reflections in parallel lines

In Figure 9.22, A(5, 2) is reflected in the line x = 3 to A′(1, 2), and A′ is then reflected in the line x = −2 to A′′(−5, 2).

A′′ is A moved 10 units left: a translation by $\begin{pmatrix} -10 \\ 0 \end{pmatrix}$. The mirrors are 5 units apart, and 10 is twice that.

Here is why it always works. Reflect P(x, y) in x = h₁, then in x = h₂:
$$P(x, y) \to P'(2h_1 - x,\ y) \to P''\big(2h_2 - (2h_1 - x),\ y\big) = P''\big(x + 2(h_2 - h_1),\ y\big)$$
Every point moves the same 2(h₂ − h₁) units sideways: a translation.

Two reflections in parallel lines make a translation, perpendicular to the lines, through twice the distance between them, in the direction from the first line towards the second. If R₁ and R₂ are reflections in x = h₁ and x = h₂ (lines parallel to the Y-axis):
$$R_2 \circ R_1 = T = 2\begin{pmatrix} h_2 - h_1 \\ 0 \end{pmatrix} \qquad R_1 \circ R_2 = T = 2\begin{pmatrix} h_1 - h_2 \\ 0 \end{pmatrix}$$
If r₁ and r₂ are reflections in y = k₁ and y = k₂ (lines parallel to the X-axis):
$$r_2 \circ r_1 = T = 2\begin{pmatrix} 0 \\ k_2 - k_1 \end{pmatrix} \qquad r_1 \circ r_2 = T = 2\begin{pmatrix} 0 \\ k_1 - k_2 \end{pmatrix}$$
In Figure 9.22: $2\begin{pmatrix} -2 - 3 \\ 0 \end{pmatrix} = \begin{pmatrix} -10 \\ 0 \end{pmatrix}$. ✓

Two rotations about the same center

In Figure 9.23, A(4, 2) is rotated −180∘° about the origin to A′(−4, −2), and A′ is rotated a further −90∘° to A′′(−2, 4).

Altogether A has turned −270∘°. The rule P(x, y) → P′(−y, x) takes A(4, 2) to (−2, 4), so it is the same as a rotation of +90∘° about O, as −270∘° and +90∘° end in the same place (they differ by a full turn, 360∘°).

Two rotations about the same center make a rotation about that center through the sum of the two angles:
$$R_2 \circ R_1 = R_1 \circ R_2 = R\big[(0, 0),\ \theta_1 + \theta_2\big]$$
for R₁[(0, 0), θ₁] and R₂[(0, 0), θ₂]. The order doesn’t matter.

Two translations

In Figure 9.24, A(2, 1) is translated by $T_1 = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ to A′(5, 2), and A′ by $T_2 = \begin{pmatrix} 1 \\ -5 \end{pmatrix}$ to A′′(6, −3).

$$A''(6, -3) = (2 + 4,\ 1 - 4), \text{ and } \begin{pmatrix} 4 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ -5 \end{pmatrix}$$

Two translations make a translation by the sum of their vectors. If $T_1 = \begin{pmatrix} a \\ b \end{pmatrix}$ and $T_2 = \begin{pmatrix} c \\ d \end{pmatrix}$,
$$T_2 \circ T_1 = T_1 + T_2 = \begin{pmatrix} a + c \\ b + d \end{pmatrix} \qquad T_1 \circ T_2 = T_2 + T_1 = \begin{pmatrix} c + a \\ d + b \end{pmatrix}$$
These are equal, so the order doesn’t matter: T₁ ∘ T₂ = T₂ ∘ T₁.

Two enlargements with the same center

In Figure 9.25, △ABC with A(1, 2), B(2, 1) and C(3, 2) is enlarged by E[(0, 0), 2] to △A′B′C′, and △A′B′C′ is enlarged again by E[(0, 0), 2] to △A′′B′′C′′.

• a.What is the center of the enlargement that takes △ABC straight to △A′′B′′C′′?
• b.What is its scale factor?

The lines AA′′, BB′′ and CC′′ all pass through O, so the center is still O(0, 0). Each enlargement doubled every length, so together they multiply it by 2 × 2 = 4: the combined enlargement is E[(0, 0), 4].
$$A(1, 2) \to A''(4 \times 1,\ 4 \times 2) = A''(4, 8) \quad B(2, 1) \to B''(8, 4) \quad C(3, 2) \to C''(12, 8)$$

Two enlargements with the same center make an enlargement with that center, whose scale factor is the product of the two:
$$E_2 \circ E_1 = E_1 \circ E_2 = E\big[(0, 0),\ k_1 k_2\big] \quad \text{for } E_1[(0, 0), k_1] \text{ and } E_2[(0, 0), k_2]$$
$$E_2 \circ E_1 = E_1 \circ E_2 = E\big[(a, b),\ k_1 k_2\big] \quad \text{for } E_1[(a, b), k_1] \text{ and } E_2[(a, b), k_2]$$

Worked examples

Question 1

Question 2

Question 3

Questions 4 to 7

9.3.2 Combining two transformations of different kinds

If F and G are two different transformations:

• a.In G ∘ F, F is done first, and its image is then transformed by G.
• b.In F ∘ G, G is done first, and its image is then transformed by F.
• c.To find the single transformation that does the same, follow a general point P(x, y) through both steps, and compare the final coordinates with the rules you know (the table in 9.1).

Worked examples

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Moving a point or a figure by working with matrices is called a matrix transformation. A point P(x, y) is written as the column $\begin{pmatrix} x \\ y \end{pmatrix}$, and a figure as one column per vertex.

9.4.1 Translation using a 2 × 1 matrix

In Figure 9.55, A(−2, 1) is translated by $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$, to A′(−2 + 5, 1 + 2) = A′(3, 3). With matrices:

$$\text{Object } O = \begin{pmatrix} -2 \\ 1 \end{pmatrix}, \quad \text{translation } T = \begin{pmatrix} 5 \\ 2 \end{pmatrix}, \quad \text{image } I = O + T = \begin{pmatrix} -2 \\ 1 \end{pmatrix} + \begin{pmatrix} 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$$

In general, for $T = \begin{pmatrix} a \\ b \end{pmatrix}$ and P(x, y):
$$\text{Image } I = \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} x + a \\ y + b \end{pmatrix}$$

Image = Object + Translation vector. A translation is done by adding a 2 × 1 matrix.

9.4.2 Transformation using a 2 × 2 matrix

The other transformations are done by multiplying by a 2 × 2 matrix. Recall from Unit 6 how a 2 × 2 matrix multiplies a column: each row of the matrix times the column, added up.
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}$$
So the matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ sends P(x, y) to P′(ax + by, cx + dy). Each new coordinate is a mixture of the old x and y.

a. The matrix for the reflection in the X-axis

The reflection in the X-axis sends P(x, y) to P′(x, −y). We want the matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ that does this. Write each new coordinate as so many x plus so many y:
x′ = x = 1 × x + 0 × y … (i)
y′ = −y = 0 × x + (−1) × y … (ii)
In matrix form, the numbers in front of x and y make the matrix:
$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$
So the reflection in the X-axis is the matrix $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. Here $\begin{pmatrix} x' \\ y' \end{pmatrix}$ is the image matrix, $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ the transformation matrix and $\begin{pmatrix} x \\ y \end{pmatrix}$ the object matrix:

Image matrix (I) = Transformation matrix (M) × Object matrix (O)

b. The matrix for the rotation of +90° about the origin

A rotation of +90∘° (anticlockwise) about the origin sends P(x, y) to P′(−y, x):
x′ = −y = 0 × x + (−1) × y … (i)
y′ = x = 1 × x + 0 × y … (ii)
$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$
So the rotation of +90∘° about the origin is $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

c. The matrix for an enlargement with center (0, 0) and scale factor k

E[(0, 0), k] sends P(x, y) to P′(kx, ky):
x′ = kx = k × x + 0 × y … (i)
y′ = ky = 0 × x + k × y … (ii)
$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$
So the enlargement E[(0, 0), k] is $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$.

Transforming a whole figure

To transform a figure, write its vertices side by side as the columns of one matrix, and multiply:

Image matrix $(I)_{2 \times n}$ = Transformation matrix $(M)_{2 \times 2}$ × Object matrix $(O)_{2 \times n}$

where n is the number of points: 2 for a line segment, 3 for a triangle, 4 for a quadrilateral. Each column of the answer is the image of the vertex in the same column.

Transformation	Object	Image	2 × 2 matrix
reflection in the X-axis	P(x, y)	P′(x, −y)	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
reflection in the Y-axis	P(x, y)	P′(−x, y)	$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
reflection in y = x (y − x = 0)	P(x, y)	P′(y, x)	$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
reflection in y = −x (x + y = 0)	P(x, y)	P′(−y, −x)	$\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$
rotation +90∘° about O (the same as −270∘°)	P(x, y)	P′(−y, x)	$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
rotation −90∘° about O (the same as +270∘°)	P(x, y)	P′(y, −x)	$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$
rotation ±180∘° about O	P(x, y)	P′(−x, −y)	$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$
rotation ±360∘° about O	P(x, y)	P′(x, y)	$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
enlargement E[(0, 0), k]	P(x, y)	P′(kx, ky)	$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$