Planar transformations AI HL

IB Mathematics: Applications and Interpretation HL · Unit 3

Lesson 3.9 · Planar transformations

Transform points and shapes with matrices, compose transformations, and interpret determinants as area scale factors.

Praise for God’s creative works

“I will praise thee, O LORD, with my whole heart; I will shew forth all thy marvellous works.” Psalm 9:1 (KJV)

Translations, reflections, and rotations can produce striking patterns from a simple starting shape. As you explore the beauty of these patterns, take a moment to thank God for the creativity and order we can observe in the world around us.

Learning objectives

By the end of this lesson, you should be able to:

use matrices to reflect, stretch, enlarge, and rotate points and shapes in two dimensions;
include a translation when applying an affine transformation;
combine transformations in the correct order;
use a determinant to find the area of a transformed shape; and
recognize how repeated transformations can generate patterns and fractals.

1. Transforming a point with a matrix

Write a point $P(x,y)$ as a column vector. A $2\times2$ matrix transforms it into a new point $P'(x',y')$ :

$\begin{pmatrix}x'\\y'\end{pmatrix} = \begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}ax+by\\cx+dy\end{pmatrix}$

To transform a polygon, apply the same matrix to each vertex, then join the image vertices in the original order.

Worked example · Image of a triangle

A triangle has vertices $A(0,0)$ , $B(2,0)$ , and $C(1,3)$ . Apply $M=\begin{pmatrix}2&0\\0&1\end{pmatrix}$ .

$A'(0,0),\qquad B'(4,0),\qquad C'(2,3)$

Every horizontal coordinate doubles; every vertical coordinate stays the same. This is a horizontal stretch with scale factor $2$ .

2. Reflection matrices

A reflection flips a shape across a line. Useful matrices include:

Line of reflection	Matrix	Coordinate rule
$x$ -axis	$\begin{pmatrix}1&0\\0&-1\end{pmatrix}$	$(x,y)\mapsto(x,-y)$
$y$ -axis	$\begin{pmatrix}-1&0\\0&1\end{pmatrix}$	$(x,y)\mapsto(-x,y)$
$y=x$	$\begin{pmatrix}0&1\\1&0\end{pmatrix}$	$(x,y)\mapsto(y,x)$
$y=-x$	$\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$	$(x,y)\mapsto(-y,-x)$

More generally, reflection in the line through the origin making angle $\theta$ with the positive $x$ -axis has matrix:

$F_\theta= \begin{pmatrix} \cos 2\theta&\sin 2\theta\\ \sin 2\theta&-\cos 2\theta \end{pmatrix}$

Worked example · Reflection in $y=x$

Reflect $P(4,-2)$ in the line $y=x$ .

$\begin{pmatrix}0&1\\1&0\end{pmatrix} \begin{pmatrix}4\\-2\end{pmatrix} = \begin{pmatrix}-2\\4\end{pmatrix}$

The image is $P'(-2,4)$ . The coordinates have exchanged places.

3. Stretches and enlargements

A horizontal stretch changes $x$ while leaving $y$ unchanged. A vertical stretch changes $y$ while leaving $x$ unchanged. An enlargement about the origin changes both coordinates by the same scale factor.

Transformation	Matrix	Effect on $(x,y)$
Horizontal stretch, factor $k$	$\begin{pmatrix}k&0\\0&1\end{pmatrix}$	$(kx,y)$
Vertical stretch, factor $k$	$\begin{pmatrix}1&0\\0&k\end{pmatrix}$	$(x,ky)$
Enlargement about $(0,0)$ , factor $k$	$\begin{pmatrix}k&0\\0&k\end{pmatrix}$	$(kx,ky)$

Worked example · Distinguishing stretch and enlargement

Transform $P(-2,3)$ using a vertical stretch of factor $3$ , then compare it with an enlargement of factor $3$ .

$\begin{pmatrix}1&0\\0&3\end{pmatrix} \begin{pmatrix}-2\\3\end{pmatrix} = \begin{pmatrix}-2\\9\end{pmatrix}$

$\begin{pmatrix}3&0\\0&3\end{pmatrix} \begin{pmatrix}-2\\3\end{pmatrix} = \begin{pmatrix}-6\\9\end{pmatrix}$

The vertical stretch gives $(-2,9)$ ; the enlargement gives $(-6,9)$ .

4. Rotation matrices

For a positive angle $\theta$ , an anticlockwise rotation about the origin uses:

$R_\theta= \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}$

A clockwise rotation through a positive angle $\theta$ uses:

$R_{-\theta}= \begin{pmatrix} \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta \end{pmatrix}$

For example, a $90^\circ$ anticlockwise rotation maps $(x,y)$ to $(-y,x)$ .

Worked example · Rotate a point

Rotate $P(3,1)$ through $90^\circ$ anticlockwise about the origin.

$\begin{pmatrix}0&-1\\1&0\end{pmatrix} \begin{pmatrix}3\\1\end{pmatrix} = \begin{pmatrix}-1\\3\end{pmatrix}$

Therefore $P'=(-1,3)$ . The distance from the origin remains $\sqrt{10}$ .

The position vector turns while its length remains unchanged.

5. Adding a translation

A $2\times2$ matrix alone sends the origin to the origin. To move the whole image after applying a matrix, add a translation vector. This gives an affine transformation:

$\begin{pmatrix}x'\\y'\end{pmatrix} = \begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} + \begin{pmatrix}e\\f\end{pmatrix}$

The matrix acts first; the vector $\begin{pmatrix}e\\f\end{pmatrix}$ is added afterward.

Worked example · Rotate, then translate

Rotate $P(2,1)$ through $90^\circ$ anticlockwise, then translate $3$ units right and $2$ units down.

$\begin{pmatrix}x'\\y'\end{pmatrix} = \begin{pmatrix}0&-1\\1&0\end{pmatrix} \begin{pmatrix}2\\1\end{pmatrix} + \begin{pmatrix}3\\-2\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix} + \begin{pmatrix}3\\-2\end{pmatrix} = \begin{pmatrix}2\\0\end{pmatrix}$

The final image is $P'(2,0)$ .

Enlargement about another centre: If the centre is the vector $\mathbf{c}$ , first subtract $\mathbf{c}$ , enlarge, and then add $\mathbf{c}$ . For scale factor $k$ , $\mathbf{x}'=k\mathbf{x}+(1-k)\mathbf{c}$ .

6. Compositions: order matters

If matrix $A$ acts first and matrix $B$ acts second, the combined matrix is $BA$ :

$\mathbf{x}\xrightarrow{A}A\mathbf{x} \xrightarrow{B}B(A\mathbf{x})=(BA)\mathbf{x}$

In general, $BA\ne AB$ . For affine transformations, keep track of the translation as well:

$\mathbf{x}\xrightarrow{\,A,\mathbf{u}\,}A\mathbf{x}+\mathbf{u} \xrightarrow{\,B,\mathbf{v}\,} BA\mathbf{x}+B\mathbf{u}+\mathbf{v}$

Worked example · Stretch, then rotate

Apply a horizontal stretch of factor $2$ to $P(1,2)$ , followed by a $90^\circ$ anticlockwise rotation.

$\begin{pmatrix}0&-1\\1&0\end{pmatrix} \begin{pmatrix}2&0\\0&1\end{pmatrix} = \begin{pmatrix}0&-1\\2&0\end{pmatrix}$

$\begin{pmatrix}0&-1\\2&0\end{pmatrix} \begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}-2\\2\end{pmatrix}$

If the rotation were performed first, the final point would instead be $(-4,1)$ .

7. Determinants and area

For $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , its determinant is:

$\det M=ad-bc$

The absolute value of the determinant is the area scale factor:

$\text{Area of image}=|\det M|\times\text{Area of object}$

If $|\det M|=1$ , area is unchanged.
If $\det M=0$ , a two-dimensional shape collapses to a line or point, so its image has zero area.
A negative determinant reverses orientation, as a reflection does. The area itself is never negative.
A translation does not change area.

Worked example · Area under a combined transformation

A shape of area $18\text{ cm}^2$ is transformed using $M=\begin{pmatrix}2&1\\0&-3\end{pmatrix}$ , then translated. Find its image area.

$\det M=(2)(-3)-(1)(0)=-6$

$\text{Image area}=|-6|\times18=\boxed{108\text{ cm}^2}$

The translation has no effect on this result.

8. Repeated transformations and fractals

Applying a transformation repeatedly is called iteration. A collection of transformations can also act on smaller copies of a shape to build a fractal-like pattern. For example, one rule might halve a shape and place it on the left; another might halve it and place it on the right.

$T_1(\mathbf{x})=\frac12\mathbf{x}, \qquad T_2(\mathbf{x})=\frac12\mathbf{x} +\begin{pmatrix}1\\0\end{pmatrix}$

Worked example · Two iterations

Start at $\mathbf{x}_0=\begin{pmatrix}2\\2\end{pmatrix}$ . Apply $T_1$ , then $T_2$ .

$\mathbf{x}_1=T_1(\mathbf{x}_0)=\begin{pmatrix}1\\1\end{pmatrix}$

$\mathbf{x}_2=T_2(\mathbf{x}_1) =\frac12\begin{pmatrix}1\\1\end{pmatrix} +\begin{pmatrix}1\\0\end{pmatrix} =\begin{pmatrix}1.5\\0.5\end{pmatrix}$

Changing the sequence of rules changes the resulting path and pattern.

IB-style Paper 1 · Skill practice

Reflect $P(-3,5)$ in the $x$ -axis. State the matrix and the image coordinates. [3 marks]

Show solution

$\begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}-3\\5\end{pmatrix} =\begin{pmatrix}-3\\-5\end{pmatrix}$

The image is $(-3,-5)$ .
Find the image of $Q(4,-1)$ under a clockwise rotation of $90^\circ$ about the origin. [3 marks]

Show solution

$\begin{pmatrix}0&1\\-1&0\end{pmatrix} \begin{pmatrix}4\\-1\end{pmatrix} =\begin{pmatrix}-1\\-4\end{pmatrix}$

The image is $(-1,-4)$ .
A rectangle has vertices $(0,0)$ , $(3,0)$ , $(3,2)$ , and $(0,2)$ . Apply a horizontal stretch with scale factor $2$ . Give its new vertices and area. [5 marks]

Show solution

The vertices become $(0,0)$ , $(6,0)$ , $(6,2)$ , and $(0,2)$ . The area changes from $6$ to $\boxed{12}$ square units.
The transformation is $\mathbf{x}'=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\mathbf{x}+\begin{pmatrix}2\\3\end{pmatrix}$ . Find the image of $(4,5)$ and describe the two steps. [4 marks]

Show solution

Reflect in the $x$ -axis to obtain $(4,-5)$ ; then translate $2$ right and $3$ up to obtain $\boxed{(6,-2)}$ .
A shape with area $15\text{ cm}^2$ is transformed by $M=\begin{pmatrix}3&1\\2&2\end{pmatrix}$ . Find the image area. [3 marks]

Show solution

$\det M=3(2)-1(2)=4$

The image area is $4(15)=\boxed{60\text{ cm}^2}$ .
A rotation of $90^\circ$ anticlockwise is followed by a reflection in the $x$ -axis. Find the combined matrix and identify the resulting single transformation. [5 marks]

Show solution

$\begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}0&-1\\1&0\end{pmatrix} = \begin{pmatrix}0&-1\\-1&0\end{pmatrix}$

This maps $(x,y)$ to $(-y,-x)$ , a reflection in $y=-x$ .

Advanced applications

Logo design. A triangular logo has vertices $A(0,0)$ , $B(4,0)$ , and $C(1,3)$ . It is rotated $90^\circ$ anticlockwise about the origin, then translated by $\begin{pmatrix}5\\2\end{pmatrix}$ . Find the three image vertices and the image area. [7 marks]

Show solution

Rotation gives $(0,0)$ , $(0,4)$ , and $(-3,1)$ . Translation gives $A'(5,2)$ , $B'(5,6)$ , and $C'(2,3)$ . The original area is $\frac12(4)(3)=6$ square units. A rotation and translation preserve area, so the image area is $\boxed{6}$ square units.
Image processing. A photograph is stretched horizontally by a factor of $1.5$ and vertically by a factor of $0.8$ . Its original area is $240\text{ cm}^2$ . Find the transformation matrix and final area. [5 marks]

Show solution

$M=\begin{pmatrix}1.5&0\\0&0.8\end{pmatrix}, \qquad \det M=1.5(0.8)=1.2$

The area is $1.2(240)=\boxed{288\text{ cm}^2}$ .
Reverse engineering. A linear transformation sends $(1,0)$ to $(2,1)$ and $(0,1)$ to $(-1,3)$ . Find its matrix, the image of $(4,-2)$ , and its area scale factor. [7 marks]

Show solution

The images of $(1,0)$ and $(0,1)$ are the columns of the matrix:

$M=\begin{pmatrix}2&-1\\1&3\end{pmatrix}$

$M\begin{pmatrix}4\\-2\end{pmatrix} =\begin{pmatrix}10\\-2\end{pmatrix}, \qquad |\det M|=|6+1|=\boxed{7}$
Two affine transformations. First transform a point with $T_1(\mathbf{x})=2\mathbf{x}+\begin{pmatrix}1\\0\end{pmatrix}$ . Then apply $T_2(\mathbf{x})=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\mathbf{x}+\begin{pmatrix}0\\3\end{pmatrix}$ . Express $T_2(T_1(\mathbf{x}))$ in the form $M\mathbf{x}+\mathbf{t}$ and find the image of $(1,2)$ . [8 marks]

Show solution

Write $R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ . Then:

$T_2(T_1(\mathbf{x})) =R\left(2\mathbf{x}+\begin{pmatrix}1\\0\end{pmatrix}\right) +\begin{pmatrix}0\\3\end{pmatrix} = \begin{pmatrix}0&-2\\2&0\end{pmatrix}\mathbf{x} +\begin{pmatrix}0\\4\end{pmatrix}$

For $(1,2)$ , the result is $\boxed{(-4,6)}$ .
Repeated pattern. Starting with $P_0(8,4)$ , repeatedly apply $T(\mathbf{x})=\frac12\mathbf{x}+\begin{pmatrix}1\\-1\end{pmatrix}$ . Find $P_1$ , $P_2$ , and the fixed point $P$ satisfying $T(P)=P$ . [7 marks]

Show solution

$P_1=(5,1),\qquad P_2=(3.5,-0.5)$

For a fixed point, $\mathbf{p}=\frac12\mathbf{p}+\begin{pmatrix}1\\-1\end{pmatrix}$ . Therefore $\frac12\mathbf{p}=\begin{pmatrix}1\\-1\end{pmatrix}$ and $\boxed{P=(2,-2)}$ .

Reflection · Recognizing patterns

A small set of transformation rules can create a surprising variety of images. Psalm 9:1 invites us to notice God’s marvellous works and respond with wholehearted praise. Which pattern or shape in nature gives you a reason to pause and praise God?

Lesson 3.9 · Planar transformations

Praise for God’s creative works

Learning objectives

1. Transforming a point with a matrix

2. Reflection matrices

3. Stretches and enlargements

4. Rotation matrices

5. Adding a translation

6. Compositions: order matters

7. Determinants and area

8. Repeated transformations and fractals

IB-style Paper 1 · Skill practice

Advanced applications

Reflection · Recognizing patterns

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Planar transformations AI HL

Lesson 3.9 · Planar transformations

Praise for God’s creative works

Learning objectives

1. Transforming a point with a matrix

2. Reflection matrices

3. Stretches and enlargements

4. Rotation matrices

5. Adding a translation

6. Compositions: order matters

7. Determinants and area

8. Repeated transformations and fractals

IB-style Paper 1 · Skill practice

Advanced applications

Reflection · Recognizing patterns

Related

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alphy school publishers

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