The matrix used to reflect a point $P(x, y)$ in the x-axis is:

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

When you look at yourself in a mirror you seem to see your body behind the mirror. Your body is in front of the mirror as your image is behind it.

An object is reflected in the mirror to form an image which is:

1. The same size as the object
2. The same distance from the mirror as the object

So reflection is an example of ISOMETRIC MAPPING.

The mirror is the line of symmetry between the object and the image.

Example 1

Find the image of the point A (2,3) after reflection in the x–axes.

Solution

Plot point A and its image A¢ such that AA¢ crosses the x–axis at B and also perpendicular to it.

For reflection AB should be the same as BA¢ i.e. $AB = BA¢$

From the figure, the coordinates of A¢ are A¢ (2,-3). So the image of A (2,3) under reflection in the x-axis is A¢ (2,-3)

Normally the letter M is used to denote reflection and thus Mx means reflection in the x–axis.

Generally $Mx(x, y) = (x, -y)$ and $My(x, y) = (-x, y)$

So $Mx(2, 3) = (2, -3)$.

Where Mx means reflection in the x–axis and My means reflection in the y-axis.

Example 2

Find the image of B(3,4) under reflection in the y–axis.

Solution

From $My(x, y) = (-x, y)$

$My(3, 4) = (-3, 4)$

Therefore the image of B(3,4) is B'(-3,4).

The line $y = x$ makes an angle $45^\circ$ with x and y axes. It is the line of symmetry for the angle YOX formed by two axis. By using isosceles triangle properties, reflection of the point (1,0) in the line $y = x$ will be (0,1) while the reflection of (0,2) in the line $y = x$ will be (2, 0) it can be noticed that the coordinates are exchanging positions. Hence the reflection of the point $(x, y)$ in the line $y = x$ is $(y, x)$.

Generally

$M_{y = x}(x, y) = (y, x)$

Where $M_{y = x}$ means reflection in the line $y = x$.

Example 3

Find the image of the point A(1,2) after reflection in the line $y = x$. Draw a sketch.

Therefore the image of A(1,2) is A'(2,1).

The reflection of the point $B(x, y)$ in the line $y = -x$ is $B'(-y, -x)$.

Example 4

Find the image of B (3,4) after reflection in the line $y = -x$ followed by another reflection in the line $y = 0$. Draw a sketch.

Solution

Reflection of B in the line $y = -x$ is B'(-4, -3). The line $y = 0$ is the x–axis. So reflection (-4, -3) in the x-axis is (-4, 3)

Therefore the image of B (3,4) is B¢(-4,3).

If the line passes through the origin and makes an angle a with x–axis in the positive direction, then its equation is $y = x \tan \alpha$ where $\tan \alpha$ is the slope of the line.

Consider the following diagram.

From the figure above OP is inclined at β and the coordinates of P are $(x, y)$.

But OPQ is a right angled triangle.

So $x = OP \cos \beta$ and $y = OP \sin \beta$.

Again OP¢R is a right angled triangle and the angle P¢QR = α - β + α - β + β, this is due to the fact that reflection is an isometric mapping.

$x' = OP' \left[ \cos(\alpha - \beta) \right] \quad \text{and} \quad y' = OP' \left[ \sin(2\alpha - \beta) \right]$

on expanding:

$x' = OP' \left[ \cos 2\alpha \cos \beta + \sin 2\alpha \sin \beta \right] \quad \ldots \ldots \ldots (1)$

$y' = OP' \left[ \sin 2\alpha \cos \beta - \sin \beta \cos 2\alpha \right] \quad \ldots \ldots \ldots (2)$

But $OP = OP'$, then:

$x' = OP \left[ \cos 2\alpha \cos \beta + \sin 2\alpha \sin \beta \right]$

$y' = OP \left[ \sin 2\alpha \cos \beta - \sin \beta \cos 2\alpha \right]$

Which implies that:

$x' = (OP \cos \beta) \cos 2\alpha + (OP \sin \beta) \sin 2\alpha$

$y' = (OP \cos \beta) \sin 2\alpha - (OP \sin \beta) \cos 2\alpha$

Remember that:

$x = OP \cos \beta \quad \text{and} \quad y = OP \sin \beta$

Therefore:

$x' = x \cos 2\alpha + y \sin 2\alpha$

$y' = x \sin 2\alpha - y \cos 2\alpha$

$M[x, y] = [x', y'] \quad \text{where}$

x' = x \cos 2\alpha + y \sin 2\alpha \\ y' = x \sin 2\alpha - y \cos 2\alpha \end{cases}$$ The above two linear equations are called **transformation equations for reflection**. These two equations can be written in matrix form as: $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$ Where: $\begin{pmatrix} x' \\ y' \end{pmatrix}$ is the position vector which is the image of $\begin{pmatrix} x \\ y \end{pmatrix}$ under reflection M, and: $\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}$ is the matrix of reflection. **NB:** $\begin{pmatrix} x \\ y \end{pmatrix} = (x, y)$ and $\begin{pmatrix} x' \\ y' \end{pmatrix} = (x', y')$ Now the angle P¢OR = 2α - β, then It follows therefore that if M is a reflection in the line inclined at α, then ### Example 5 Find the image of the point A (1, 2) after a reflection in the line $y = x$. **Solution** The line $y = x$ has a slope 1 So $\tan \alpha = 1$, $\alpha = 45^\circ$ <img alt="" src="/media/assets/legacy/2023/11/44_1445338683636-e1745973750105.png" width="304" height="265"/> 45° From $M_\alpha = \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix}$ Then $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos 90^\circ & \sin 90^\circ \\ \sin 90^\circ & -\cos 90^\circ \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$, but $(x, y) = (1, 2)$ So $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 + 2 \\ 1 + 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ Therefore $(x', y') = (2, 1)$ which is the image of (1, 2). ### Example 6 Find the image of B (3,4) after reflection in the line $y = -x$ followed by another reflection in the line $y = 0$. **Solution** The line $y = -x$ has slope -1 So $\tan \alpha = -1$ $\alpha = -45^\circ = 135^\circ$ $$M_\alpha = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$ But $(x, y) = (3, 4)$ $$M_{135^\circ} = \begin{bmatrix} \cos 270^\circ & \sin 270^\circ \\ \sin 270^\circ & -\cos 270^\circ \end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 0 + -4 \\ -3 + 0 \end{bmatrix} = \begin{bmatrix} -4 \\ -3 \end{bmatrix}$$ Now we need the image of $(-4, -3)$ after being reflected in the line $y = 0$. But the line $y = 0$ has 0 slope because it is the x–axis, ### Example 7 Find the equation of the line $y = 2x + 5$ after being reflected in the line $y = x$, **Solution** The line $y = x$ has a slope 1 So $\tan \alpha = 1$ which means $\alpha = 45^\circ$ To find the image of the line $y = 2x + 5$, we choose at least two points on it and find their images, then we use the image points to find the equation of the image line. | x | 0 | 1 | |---|---|---| | y | 5 | 7 | But $M_\alpha = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ So $2\alpha = 2 \times 45^\circ = 90^\circ$ For $(0,5)$, $\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos 90^\circ & \sin 90^\circ \\ \sin 90^\circ & -\cos 90^\circ \end{bmatrix} \begin{bmatrix} 0 \\ 5 \end{bmatrix}$ $= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 5 \end{bmatrix} = \begin{bmatrix} 0+5 \\ 0+0 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix}$ For $(1,7)$, $\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos 90^\circ & \sin 90^\circ \\ \sin 90^\circ & -\cos 90^\circ \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix}$ $= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix} = \begin{bmatrix} 0+7 \\ 1+0 \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$ Slope $= \frac{1-0}{7-5} = \frac{1}{2}$ Equation: $\frac{y-0}{x-5} = \frac{1}{2}$ $2y = x - 5$ $y = \frac{x-5}{2}$ Therefore, the equation of the line $y = 2x + 5$ after being reflected in the line $y = x$ is $2y = x - 5$. Now $y = 2x + 5$ The points (0,5) and (1,7) lie on the line So the image line is the line passing through (5,0) and (7,1) and it is obtained as follows;

Definition

A rotation is a transformation which moves a point through a given angle about a fixed point.

The point $P(x, y)$ is rotated through angle $\theta$ to form an image $P'(x', y')$.

Rotation is an isometric mapping and it is usually denoted by R.

Therefore $R_\theta$ means rotation of an object through an angle θ.

In the xy plane, when θ is measured in the clockwise direction it is negative and when it is measured in the anticlockwise direction it is positive.

(a) Zero matrix

A zero matrix is the matrix whose elements are all zeros.

e.g.

$A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \quad B = \begin{pmatrix} 0 & 0 \end{pmatrix} \quad C = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad \text{or} \quad D = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

(b) Square matrix

A square matrix is a matrix whose number of rows is equal to the number of columns.

For example

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \quad \text{or} \quad B = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 6 & 8 \\ 5 & 7 & 9 \end{pmatrix}$

(c) Identity Matrix

A square matrix whose elements in the leading diagonal are ones and all other elements are zeros.

e.g.

$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ or $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

(d) Column Matrix

A matrix with only one column.

e.g.

$\mathbf{A} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ or $\mathbf{B} = \begin{pmatrix} 1 \\ 6 \\ 9 \end{pmatrix}$

(e) Row Matrix

A matrix with only one row.

e.g.

$\mathbf{D} = \begin{pmatrix} 1 & 10 \end{pmatrix}$ or $\mathbf{E} = \begin{pmatrix} 3 & -2 & 17 \end{pmatrix}$