Introduction to Transformations
Hello dear student! Welcome to Unit 6 of your Grade 11 Mathematics. In this unit, we are going to study Transformations of the Plane. Have you ever looked at your reflection in a mirror? Or watched a wheel spinning? Or used a photocopier to make a bigger copy of a drawing? All of these are examples of transformations!
A transformation is a rule or operation that changes the position, shape, or size of a figure in a plane. We start with an original figure (called the object or pre-image) and apply a rule to get a new figure (called the image).
Throughout this unit, we will use the notation $P’$ (read as “P prime”) to represent the image of point $P$ under a transformation. The transformation itself is often denoted by the letter $T$, so we write:
This means: “Transformation $T$ maps point $P$ to point $P’$.”
Types of Transformations
In Grade 11, we study four main types of transformations:
| Transformation | What it does | Preserves Shape? | Preserves Size? |
|---|---|---|---|
| Translation | Slides (shifts) the figure | Yes | Yes |
| Reflection | Flips (mirrors) the figure | Yes | Yes |
| Rotation | Turns the figure around a point | Yes | Yes |
| Enlargement | Stretches or shrinks the figure | Yes (similar) | No (unless $k=1$) |
Translation, reflection, and rotation are called rigid transformations (or isometries) because they preserve both shape AND size. Enlargement is NOT rigid because it changes the size.
Now let us study each one in detail, step by step!
Translation
What is a Translation?
A translation moves every point of a figure by the same distance, in the same direction. Think of sliding a book across a table — every point on the book moves the same amount. The shape does NOT change, the size does NOT change, and the orientation (which way it faces) does NOT change.
Translation Vector
A translation is described by a translation vector written as a column vector:
This means: move $a$ units in the x-direction (right if positive, left if negative) and $b$ units in the y-direction (up if positive, down if negative).
Formula for Translation
If point $P(x, y)$ is translated by the vector $\begin{pmatrix} a \\ b \end{pmatrix}$, then the image $P'(x’, y’)$ is:
In short: $P'(x+a,\; y+b)$
Matrix Form of Translation
We can write the translation using matrices. We use homogeneous coordinates (adding an extra row of 1s):
The $3 \times 3$ matrix $\begin{pmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$ is called the translation matrix.
Worked Example 1
Find the image of point $A(3, 4)$ under the translation by vector $\begin{pmatrix} 5 \\ -2 \end{pmatrix}$.
Solution:
Worked Example 2
Triangle $ABC$ has vertices $A(1, 2)$, $B(3, 5)$, $C(6, 1)$. Find the image of the triangle under translation by $\begin{pmatrix} -3 \\ 4 \end{pmatrix}$.
Solution:
We translate each vertex:
The image triangle is $A’B’C’$ with vertices $(-2, 6)$, $(0, 9)$, $(3, 5)$.
Worked Example 3
The point $P'(7, -1)$ is the image of $P(2, 3)$ under a translation. Find the translation vector.
Solution:
The translation vector is: $P’ – P = \begin{pmatrix} 7-2 \\ -1-3 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}$
Finding the Image of a Line Under Translation
If a line has equation $ax + by + c = 0$, and we translate by $\begin{pmatrix} p \\ q \end{pmatrix}$, then every point $(x, y)$ on the line moves to $(x+p, y+q)$. The image line is obtained by replacing $x$ with $x-p$ and $y$ with $y-q$:
Worked Example 4
Find the image of the line $2x + 3y – 6 = 0$ under the translation $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$.
Solution:
Replace $x$ with $x – 1$ and $y$ with $y – (-2) = y + 2$:
Answer:
Answer:
If $Q'(x’, y’)$ is the image of $Q(x, y)$, then $x’ = x + a$ and $y’ = y + b$. So $x = x’ – a$ and $y = y’ – b$:
Tip: To find the original point, SUBTRACT the translation vector from the image.
Answer:
Replace $x$ with $x – 3$ (and $y$ stays as $y – 0 = y$):
Reflection
What is a Reflection?
A reflection flips a figure over a line called the mirror line (or line of reflection). The image is a mirror image of the original. Think of looking at yourself in a mirror — your left hand appears as a right hand in the reflection!
Key property: The mirror line is the perpendicular bisector of the segment joining each point and its image.
Reflection in the x-axis
When we reflect a point $P(x, y)$ in the x-axis, the x-coordinate stays the same and the y-coordinate changes sign:
Reflection in the y-axis
Reflection in the line $y = x$
When we reflect in the line $y = x$, the x and y coordinates are swapped:
Reflection in the line $y = -x$
Reflection in the line $x = a$
Reflection in the line $y = b$
Reflection in the origin
| Reflection in | $P(x,y)$ maps to | Matrix |
|---|---|---|
| x-axis | $(x, -y)$ | $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ |
| y-axis | $(-x, y)$ | $\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ |
| $y = x$ | $(y, x)$ | $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ |
| $y = -x$ | $(-y, -x)$ | $\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ |
| Origin | $(-x, -y)$ | $\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ |
| $x = a$ | $(2a-x, y)$ | — |
| $y = b$ | $(x, 2b-y)$ | — |
Worked Example 5
Find the image of triangle $ABC$ with vertices $A(2, 3)$, $B(5, 1)$, $C(1, -2)$ under reflection in the x-axis.
Solution:
Reflect each vertex: $P(x,y) \to P'(x, -y)$
Worked Example 6
Find the image of point $P(4, -3)$ under reflection in the line $y = x$.
Solution:
Swap the coordinates: $P'( -3, 4)$
Worked Example 7
Find the image of point $Q(3, 5)$ under reflection in the line $x = 2$.
Solution:
Use the formula: $P'(2a – x, y) = (2(2) – 3, 5) = (1, 5)$
Check: The midpoint of $Q$ and $Q’$ is $\left(\frac{3+1}{2}, \frac{5+5}{2}\right) = (2, 5)$. The x-coordinate is 2, which lies on the line $x = 2$. ✓
Worked Example 8
Find the image of the line $y = 3x + 2$ under reflection in the y-axis.
Solution:
Replace $x$ with $-x$: $y = 3(-x) + 2 = -3x + 2$
Worked Example 9
Find the image of the line $2x – y + 3 = 0$ under reflection in the line $y = x$.
Solution:
Swap $x$ and $y$: $2y – x + 3 = 0$, which can be rewritten as:
Answer:
Formula: $P(x,y) \to P'(-y, -x)$
Answer:
Replace $y$ with $-y$:
Answer:
For reflection in $y = b$: $P'(x, 2b – y)$. Here $b = 2$, so $P'(x, 4 – y)$.
We know $P’ = (5, 3)$, so $x = 5$ and $4 – y = 3 \Rightarrow y = 1$.
Check: Midpoint of $P$ and $P’$ is $(5, 2)$, which lies on $y = 2$. ✓
Rotation
What is a Rotation?
A rotation turns a figure about a fixed point called the centre of rotation through a specified angle. The figure stays the same shape and size — it just turns around!
Have you ever turned a key in a lock? Or watched a clock’s hands move? Those are rotations!
Direction of Rotation
In mathematics, we use the following convention:
• Positive angle = Anti-clockwise (counter-clockwise) rotation
• Negative angle = Clockwise rotation
Rotation About the Origin
When we rotate point $P(x, y)$ about the origin through angle $\theta$ (anti-clockwise):
In matrix form:
The matrix $R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ is called the rotation matrix.
Standard Rotations (About the Origin)
| Rotation | $P(x,y)$ maps to | Matrix |
|---|---|---|
| $90°$ anti-clockwise | $(-y, x)$ | $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ |
| $90°$ clockwise ($-90°$) | $(y, -x)$ | $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ |
| $180°$ (either direction) | $(-x, -y)$ | $\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ |
| $270°$ anti-clockwise ($= -90°$) | $(y, -x)$ | $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ |
| $360°$ (full turn) | $(x, y)$ | $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ |
Worked Example 10
Find the image of $A(3, -2)$ under a rotation of $90°$ anti-clockwise about the origin.
Solution:
$P(x,y) \to P'(-y, x)$
Using the matrix method:
Worked Example 11
Find the image of $B(-1, 4)$ under a rotation of $180°$ about the origin.
Solution:
Worked Example 12
Find the image of $C(2, 5)$ under a rotation of $90°$ clockwise about the origin.
Solution:
$90°$ clockwise: $P(x,y) \to P'(y, -x)$
Worked Example 13
Find the image of the line $y = 2x$ under a rotation of $90°$ anti-clockwise about the origin.
Solution:
Under $90°$ anti-clockwise: $x \to -y$ and $y \to x$. So replace $x$ with $-y$ and $y$ with $x$:
Rotation About a Point Other Than the Origin
To rotate about a point $C(a, b)$ through angle $\theta$:
Step 1: Translate so that $C$ moves to the origin: $(x-a, y-b)$
Step 2: Apply the rotation about the origin
Step 3: Translate back: add $(a, b)$ to the result
Worked Example 14
Find the image of $P(4, 3)$ under a rotation of $90°$ anti-clockwise about the point $C(1, 2)$.
Solution:
Step 1: Translate $C$ to origin: $P$ becomes $(4-1, 3-2) = (3, 1)$
Step 2: Rotate $90°$ anti-clockwise: $(3, 1) \to (-1, 3)$
Step 3: Translate back: $(-1+1, 3+2) = (0, 5)$
Answer:
$270°$ anti-clockwise $= 90°$ clockwise, so $P(x,y) \to P'(y, -x)$
Answer:
Answer:
Step 1: Translate: $(5-2, 1-3) = (3, -2)$
Step 2: Rotate $90°$ anti-clockwise: $(3, -2) \to (2, 3)$
Step 3: Translate back: $(2+2, 3+3) = (4, 6)$
Enlargement (Dilation)
What is an Enlargement?
An enlargement (also called a dilation or scaling) changes the size of a figure by a factor called the scale factor ($k$). The figure is enlarged or shrunk from a fixed point called the centre of enlargement.
Unlike the other three transformations, enlargement is NOT a rigid transformation — it changes the size of the figure (unless $k = 1$ or $k = -1$).
Properties of Enlargement
• If $k > 1$: The image is bigger than the object
• If $0 < k < 1$: The image is smaller than the object (reduction)
• If $k = 1$: The image is the same as the object (identity)
• If $k < 0$: The image is on the opposite side of the centre AND the size changes by $|k|$
• If $k = -1$: Same size but on the opposite side (equivalent to reflection in the centre)
• If $k = 0$: The image is a single point (the centre itself)
Enlargement About the Origin
If point $P(x, y)$ is enlarged about the origin with scale factor $k$:
This is a scalar matrix — $k$ times the identity matrix.
Enlargement About a Point $C(a, b)$
In other words: first find the vector from $C$ to $P$, multiply it by $k$, then add it back to $C$.
Worked Example 15
Find the image of triangle $ABC$ with $A(2, 1)$, $B(4, 3)$, $C(6, 0)$ under an enlargement with scale factor $2$ about the origin.
Solution:
Worked Example 16
Find the image of $P(5, 3)$ under an enlargement with scale factor $-2$ about the origin.
Solution:
The image is on the opposite side of the origin and twice as far away.
Worked Example 17
Find the image of $Q(4, 7)$ under an enlargement with scale factor $3$ about the point $C(1, 2)$.
Solution:
Worked Example 18
Find the image of the line $y = \dfrac{1}{2}x$ under an enlargement with scale factor $3$ about the origin.
Solution:
Replace $x$ with $\dfrac{x}{3}$ and $y$ with $\dfrac{y}{3}$:
The image is the same line! Why? Because the line passes through the origin (the centre of enlargement), and any line through the centre maps to itself under enlargement.
Effect of Enlargement on Length and Area
If the scale factor is $k$:
Answer:
Answer:
Since $k = \frac{1}{2} < 1$, the image is closer to the centre — this is a reduction.
Answer:
Given $C(1, 1)$, $A(3, 5)$, $A'(9, 11)$:
From x-coordinate: $1 + 2k = 9 \Rightarrow 2k = 8 \Rightarrow k = 4$
From y-coordinate: $1 + 4k = 11 \Rightarrow 4k = 10 \Rightarrow k = 2.5$
Wait! We get different values of $k$ from the two coordinates, which means the given data is inconsistent. Let me correct the question — if $C = (1, 1)$ and $A'(9, 11)$, then we need $A’$ to satisfy both equations with the same $k$.
For consistency, if $k = 4$: $A’ = (1 + 8, 1 + 16) = (9, 17)$. So the correct image should be $A'(9, 17)$ for $k = 4$, or the centre might be different. This shows the importance of checking consistency!
Matrix Representation of Transformations
One of the most powerful tools in this unit is using matrices to represent transformations. Every linear transformation (reflection, rotation, enlargement about the origin) can be written as a $2 \times 2$ matrix.
How to Use the Transformation Matrix
If $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is the transformation matrix, then:
Worked Example 19
A transformation is represented by the matrix $M = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Find the image of $(3, -4)$ and identify the transformation.
Solution:
Image: $(4, 3)$. This is the matrix for $90°$ anti-clockwise rotation about the origin.
Finding the Transformation Matrix From Images
If we know the images of two points under an unknown transformation, we can find the matrix!
Worked Example 20
A transformation maps $(1, 0) \to (0, 1)$ and $(0, 1) \to (-1, 0)$. Find the transformation matrix and identify the transformation.
Solution:
The columns of the transformation matrix are the images of $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$:
This is the matrix for rotation of $90°$ anti-clockwise about the origin.
Worked Example 21
A transformation maps $(1, 0) \to (3, 0)$ and $(0, 1) \to (0, 3)$. Find the matrix and identify the transformation.
Solution:
This is enlargement with scale factor $3$ about the origin.
Determinant and Transformation Properties
The determinant of a transformation matrix tells us important information:
| Value of $\det(M)$ | Meaning |
|---|---|
| $\det(M) = 1$ | Area is preserved (e.g., rotation, translation, reflection in a line) |
| $\det(M) = -1$ | Area preserved but orientation is reversed (e.g., reflection) |
| $\det(M) = k^2$ (positive) | Area multiplied by $k^2$; orientation preserved (e.g., enlargement with $k > 0$) |
| $\det(M) < 0$ | Orientation is reversed |
| $\det(M) = 0$ | The transformation collapses the plane to a line or point (singular) |
Worked Example 22
Find the determinant of the reflection matrix in the y-axis and interpret it.
Solution:
Since $\det(M) = -1$: area is preserved, but orientation is reversed. This makes sense — a reflection flips the figure!
Composite Transformations
What is a Composite Transformation?
A composite transformation is two or more transformations applied one after the other. If transformation $T_1$ is applied first, and then $T_2$ is applied to the result, the overall transformation is written as $T_2 \circ T_1$ (read as “$T_2$ after $T_1$”).
Matrix Multiplication for Composites
If $T_1$ has matrix $M_1$ and $T_2$ has matrix $M_2$, then the composite $T_2 \circ T_1$ has matrix:
Important: We multiply in the REVERSE order of application! $T_1$ is applied first, so $M_1$ goes on the RIGHT.
Worked Example 23
Find the matrix representing a reflection in the x-axis followed by a reflection in the y-axis.
Solution:
Reflection in x-axis: $M_1 = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$
Reflection in y-axis: $M_2 = \begin{pmatrix}-1&0\\0&1\end{pmatrix}$
This is the matrix for reflection in the origin (or rotation of $180°$). So reflecting in the x-axis then the y-axis is the same as reflecting in the origin!
Worked Example 24
Find the single matrix equivalent to a rotation of $90°$ anti-clockwise followed by an enlargement with scale factor $2$ about the origin.
Solution:
Rotation $90°$ anti-clockwise: $M_1 = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$
Enlargement $k=2$: $M_2 = \begin{pmatrix}2&0\\0&2\end{pmatrix}$
Verification: Apply to point $(1, 0)$: $90°$ rotation gives $(0, 1)$, then enlargement gives $(0, 2)$. Using matrix: $\begin{pmatrix}0&-2\\2&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}0\\2\end{pmatrix}$ ✓
Worked Example 25
Find the image of $(3, 1)$ under: reflection in the line $y = x$ followed by reflection in the x-axis.
Solution:
Method 1 (Step by step):
Reflection in $y = x$: $(3, 1) \to (1, 3)$
Reflection in x-axis: $(1, 3) \to (1, -3)$
Method 2 (Matrix):
Inverse Transformations
The inverse of a transformation $T$ is the transformation that “undoes” $T$. It maps every image back to its original object.
If $T(P) = P’$, then $T^{-1}(P’) = P$.
Finding the Inverse Matrix
For a $2 \times 2$ matrix $M = \begin{pmatrix}a&b\\c&d\end{pmatrix}$ with $\det(M) \neq 0$:
Inverses of Standard Transformations
| Transformation | Inverse |
|---|---|
| Translation by $\begin{pmatrix}a\\b\end{pmatrix}$ | Translation by $\begin{pmatrix}-a\\-b\end{pmatrix}$ |
| Rotation by $\theta$ anti-clockwise | Rotation by $\theta$ clockwise (or $-\theta$) |
| Reflection in line $l$ | Reflection in line $l$ (itself!) |
| Enlargement with scale factor $k$ | Enlargement with scale factor $\dfrac{1}{k}$ |
Worked Example 26
Find the inverse of the rotation matrix for $90°$ anti-clockwise: $M = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$.
Solution:
This is the matrix for $90°$ clockwise rotation, which is the correct inverse! ✓
Worked Example 27
Verify that the reflection in the x-axis is its own inverse.
Solution:
Finding the Equation of the Image Line
This is a very common type of exam question. Here is the general approach:
Method 1: Using Two Points
Step 1: Find two points on the original line.
Step 2: Find the images of those two points under the transformation.
Step 3: Find the equation of the line through the two image points.
Method 2: Direct Substitution
Replace $x$ and $y$ in the equation with their expressions in terms of $x’$ and $y’$.
Worked Example 28
Find the image of the line $3x + y – 5 = 0$ under the transformation represented by $M = \begin{pmatrix}2&0\\0&3\end{pmatrix}$.
Solution using Method 2:
The transformation gives $x’ = 2x$ and $y’ = 3y$, so $x = \dfrac{x’}{2}$ and $y = \dfrac{y’}{3}$.
Substitute into the equation:
Multiply by 6:
Dropping the primes: $9x + 2y = 30$
Worked Example 29
Find the image of the line $x – 2y + 4 = 0$ under a rotation of $90°$ anti-clockwise about the origin.
Solution using Method 1:
Two points on the line: when $x = 0$: $y = 2$, so $(0, 2)$. When $y = 0$: $x = -4$, so $(-4, 0)$.
Rotate $90°$ anti-clockwise: $(0, 2) \to (-2, 0)$ and $(-4, 0) \to (0, -4)$.
Line through $(-2, 0)$ and $(0, -4)$: slope $= \dfrac{-4-0}{0-(-2)} = -2$
Properties Preserved by Transformations
| Property | Translation | Reflection | Rotation | Enlargement |
|---|---|---|---|---|
| Shape | ✓ | ✓ | ✓ | ✓ (similar) |
| Size | ✓ | ✓ | ✓ | ✗ |
| Angles | ✓ | ✓ | ✓ | ✓ |
| Parallel lines | ✓ | ✓ | ✓ | ✓ |
| Orientation (handedness) | ✓ | ✗ | ✓ | ✓ if $k>0$, ✗ if $k<0$ |
| Midpoints | ✓ | ✓ | ✓ | ✓ |
| Collinearity | ✓ | ✓ | ✓ | ✓ |
What does “orientation” mean? Imagine a triangle labeled ABC going anti-clockwise. After a reflection, the image $A’B’C’$ goes clockwise. The “handedness” has flipped — like your left hand becoming a right hand in a mirror. Translations and rotations preserve orientation; reflections reverse it.
Describing a Transformation Fully
In exam questions, you may be asked to “fully describe” a transformation. Here is what you need to state for each type:
Translation: State the translation vector.
Reflection: State the mirror line (equation of the line).
Rotation: State the centre of rotation, the angle, and the direction (clockwise/anti-clockwise).
Enlargement: State the centre of enlargement and the scale factor.
Worked Example 30
$A(1, 2)$ maps to $A'(4, 5)$, $B(3, 1)$ maps to $B'(6, 4)$. Describe the transformation fully.
Solution:
Check translation vector: $A’ – A = (3, 3)$ and $B’ – B = (3, 3)$. Same vector!
Worked Example 31
$P(2, 3)$ maps to $P'(2, -3)$ and $Q(5, 1)$ maps to $Q'(5, -1)$. Describe the transformation fully.
Solution:
The x-coordinates stay the same, y-coordinates change sign. This is a reflection in the x-axis.
Verification: midpoint of $P$ and $P’$ is $(2, 0)$, which lies on the x-axis. ✓
Worked Example 32
$R(1, 0)$ maps to $R'(0, 1)$ and $S(0, 1)$ maps to $S'(-1, 0)$. Describe the transformation fully.
Solution:
Check if it is a $90°$ anti-clockwise rotation: $(1, 0) \to (0, 1)$ ✓ and $(0, 1) \to (-1, 0)$ ✓
Finding the Centre and Angle of Rotation
When given an object and its image under an unknown rotation, how do you find the centre and angle?
Finding the Centre of Rotation
Perpendicular Bisector Method:
Step 1: Join each point to its image (e.g., $AA’$, $BB’$).
Step 2: Draw the perpendicular bisector of $AA’$ and the perpendicular bisector of $BB’$.
Step 3: The intersection of these perpendicular bisectors is the centre of rotation.
Finding the Angle of Rotation
Step 1: Draw a line from the centre to a point on the object.
Step 2: Draw a line from the centre to the image of that point.
Step 3: Measure the angle between these two lines (anti-clockwise from object to image).
Algebraic Method for Finding Centre
If the centre is $C(a, b)$ and $P(x_1, y_1) \to P'(x_1′, y_1′)$ under rotation, then the distances from $C$ must be equal:
Use two such equations to solve for $a$ and $b$.
Worked Example 33
$A(2, 1)$ maps to $A'(1, -2)$ and $B(5, 3)$ maps to $B'(6, 0)$. Find the centre and angle of rotation.
Solution:
Let centre be $C(a, b)$. From equal distances for $A$:
From equal distances for $B$:
From (i) and (ii): $-3b – 3b = 1 \Rightarrow -6b = 1 \Rightarrow b = -\dfrac{1}{6}$
Finding the Mirror Line of a Reflection
Given an object point and its image under reflection, the mirror line is the perpendicular bisector of the segment joining the point and its image.
Worked Example 34
$A(3, 1)$ maps to $A'(7, 5)$ under a reflection. Find the equation of the mirror line.
Solution:
Step 1: Find the midpoint of $AA’$:
Step 2: Find the slope of $AA’$:
Step 3: The mirror line is perpendicular to $AA’$, so its slope is $-1$.
Step 4: Equation of line through $(5, 3)$ with slope $-1$:
Answer:
Finding the Centre and Scale Factor of Enlargement
Given an object point and its image under enlargement:
Step 1: Draw a line from the object point through the image point. The centre lies on this line.
Step 2: Repeat for another point-image pair. The intersection gives the centre.
Algebraically: If $C(a,b)$ is the centre, $P(x,y) \to P'(x’, y’)$ with scale factor $k$:
Use two point-image pairs to find $a$, $b$, and $k$.
Worked Example 35
$A(1, 2)$ maps to $A'(4, 5)$ and $B(3, 1)$ maps to $B'(9, -1)$ under an enlargement. Find the centre and scale factor.
Solution:
From $A \to A’$: $4 = a + k(1-a)$ and $5 = b + k(2-b)$ … (i)
From $B \to B’$: $9 = a + k(3-a)$ and $-1 = b + k(1-b)$ … (ii)
From the x-equations: $4 = a + k – ka$ and $9 = a + 3k – ka$
From $4 = a + \frac{5}{2} – \frac{5}{2}a$:
From $5 = b + \frac{5}{2}(2-b) = b + 5 – \frac{5b}{2}$:
Special Results and Shortcuts
Result 1: Reflection in the x-axis followed by reflection in the y-axis $=$ Reflection in the origin $=$ Rotation of $180°$ about the origin.
Result 2: Two successive reflections in parallel lines $=$ Translation by twice the distance between the lines (perpendicular to the lines).
Result 3: Two successive reflections in intersecting lines $=$ Rotation about the intersection point by twice the angle between the lines.
Result 4: Any isometry (rigid transformation) can be expressed as either a single translation, a single rotation, a single reflection, or a glide reflection.
Result 5: The composition of two rotations about the same centre is a single rotation (add the angles).
Worked Example 36
Two parallel mirror lines are $y = 1$ and $y = 4$. What single transformation is equivalent to reflection in $y = 1$ followed by reflection in $y = 4$?
Solution:
The distance between the lines is $4 – 1 = 3$ units.
Equivalent transformation: Translation by $2 \times 3 = 6$ units in the direction perpendicular to the lines (i.e., in the y-direction, from $y=1$ towards $y=4$, which is upward).
Worked Example 37
Two mirror lines $y = x$ and $y = -x$ intersect at the origin with an angle of $90°$ between them. What single transformation is equivalent to reflection in $y = x$ followed by reflection in $y = -x$?
Solution:
The angle between the lines is $90°$, so the equivalent rotation is $2 \times 90° = 180°$ about the origin.
Verification using matrices:
Unit Summary — Key Exam Notes
• Translation: $P(x,y) \to P'(x+a, y+b)$; matrix uses $3 \times 3$ homogeneous form
• Reflection: Learn the mapping for x-axis, y-axis, $y=x$, $y=-x$, origin, $x=a$, $y=b$
• Rotation: $90°$ ACW: $(-y,x)$; $90°$ CW: $(y,-x)$; $180°$: $(-x,-y)$; General: use matrix with $\cos\theta$, $\sin\theta$
• Enlargement: About origin: $P'(kx, ky)$; Area scale factor $= k^2$; Negative $k$ means opposite side
• Composite: Matrix $= M_2 \times M_1$ (reverse order!)
• Inverse: Reflection is self-inverse; Rotation inverse is opposite angle; Enlargement inverse uses $1/k$
• Image of line: Either transform two points, or substitute $x, y$ in terms of $x’, y’$
• Finding mirror line: Perpendicular bisector of point-image segment
• Two reflections: Parallel lines $\to$ translation; Intersecting lines $\to$ rotation
Quick Revision Notes — Transformations of the Plane
Important Definitions
Transformation: A rule that maps every point in the plane to another point in the plane.
Object (Pre-image): The original figure before transformation.
Image: The resulting figure after transformation.
Isometry (Rigid Transformation): A transformation that preserves distances (and therefore shape and size). Translation, reflection, and rotation are isometries.
Translation Vector: A column vector $\begin{pmatrix}a\\b\end{pmatrix}$ that describes the distance and direction of a translation.
Mirror Line (Line of Reflection): The line over which a figure is reflected. It is the perpendicular bisector of each point-image segment.
Centre of Rotation: The fixed point about which a figure is rotated.
Centre of Enlargement: The fixed point from which a figure is enlarged or reduced.
Scale Factor ($k$): The ratio of image length to object length in an enlargement.
Composite Transformation: Two or more transformations applied in succession.
Key Formulas — Must Memorize
Translation
Reflection Matrices
Rotation Matrices (About Origin)
Enlargement
Composite Transformations
Special Results
Determinant Quick Guide
• $\det = +1$: Isometry, orientation preserved (rotation, translation)
• $\det = -1$: Isometry, orientation reversed (reflection)
• $\det = k^2 > 0$: Enlargement with scale factor $k$, orientation preserved
• $\det = -k^2 < 0$: Enlargement with negative scale factor $|k|$, orientation reversed
• $\det = 0$: Singular transformation (collapses dimension)
Common Mistakes to Avoid
❌ Mistake 1: Getting the order wrong in composite transformations.
✅ Correct: If $T_1$ is applied first, then $T_2$, the matrix is $M_2 \times M_1$. The SECOND transformation goes on the LEFT.
❌ Mistake 2: Confusing $90°$ clockwise and $90°$ anti-clockwise.
✅ Correct: $90°$ ACW: $(x,y) \to (-y,x)$. $90°$ CW: $(x,y) \to (y,-x)$. Memorize both!
❌ Mistake 3: Using area scale factor $k$ instead of $k^2$ for enlargement.
✅ Correct: If length scale factor is $k$, area scale factor is $k^2$. Always square it!
❌ Mistake 4: Forgetting to translate back when rotating about a non-origin point.
✅ Correct: Always do THREE steps: translate to origin → rotate → translate back.
❌ Mistake 5: When finding the image of a line under reflection in $y = x$, adding something instead of just swapping.
✅ Correct: Reflection in $y = x$: simply swap $x$ and $y$ in the equation. Nothing else changes.
❌ Mistake 6: Confusing reflection in the origin with reflection in the x-axis or y-axis.
✅ Correct: Reflection in the origin: $(x,y) \to (-x,-y)$. This changes BOTH coordinates, not just one.
❌ Mistake 7: Using the wrong sign for the translation vector when finding the original point from the image.
✅ Correct: Original $=$ Image $-$ Translation vector. If image is $(7,-1)$ and vector is $(5,-4)$, original $= (2,3)$.
Quick Examples for Revision
Q: What is the image of $(4, -3)$ under reflection in the x-axis?
A: $(4, 3)$. Only the y-coordinate changes sign.
Q: What matrix represents an enlargement with scale factor $-1$ about the origin?
A: $\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$. This is the same as reflection in the origin (or $180°$ rotation).
Q: Is a reflection an isometry? Why or why not?
A: Yes! An isometry preserves distances. In a reflection, every point and its image are equidistant from the mirror line, and all distances between points are preserved. The shape and size don’t change — only the orientation flips.
Q: What single transformation is equivalent to reflecting in $y = 2$ then reflecting in $y = 5$?
A: Translation by $\begin{pmatrix}0\\6\end{pmatrix}$. The lines are parallel, distance $= 5-2 = 3$, so translation $= 2 \times 3 = 6$ units upward (from first line towards second line).
Q: Find $\det\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ and interpret.
A: $\det = 0 \times 0 – (-1) \times 1 = 1$. Since $\det = +1$, this is an isometry with orientation preserved — consistent with rotation of $90°$ anti-clockwise.
Challenge Exam Questions — Transformations of the Plane
Test yourself with these difficult questions. Try each one before looking at the answer!
Multiple Choice Questions
Q1. The point $(2, -5)$ is reflected in the line $y = -x$. The image is:
(a) $(5, -2)$ (b) $(-5, 2)$ (c) $(5, 2)$ (d) $(-5, -2)$
Answer: (a) $(5, -2)$
Reflection in $y = -x$: $P(x,y) \to P'(-y, -x)$
Q2. The matrix $\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$ represents:
(a) Rotation of $90°$ anti-clockwise (b) Rotation of $90°$ clockwise (c) Reflection in $y = x$ (d) Reflection in $y = -x$
Answer: (b) Rotation of $90°$ clockwise
For $90°$ clockwise: $(x,y) \to (y,-x)$, giving matrix $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. ✓
Verification: $\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}0\\-1\end{pmatrix}$, and $(1,0) \to (0,-1)$ under $90°$ CW. ✓
Q3. Under an enlargement with scale factor $k$, the area of a shape changes by a factor of:
(a) $k$ (b) $2k$ (c) $k^2$ (d) $\sqrt{k}$
Answer: (c) $k^2$
Area is a two-dimensional measure. If all lengths are multiplied by $k$, area is multiplied by $k \times k = k^2$.
Q4. The composition of a reflection in the x-axis followed by a reflection in the line $y = x$ is equivalent to:
(a) Rotation of $90°$ ACW (b) Rotation of $90°$ CW (c) Rotation of $180°$ (d) Reflection in $y = -x$
Answer: (b) Rotation of $90°$ CW
Wait — $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is $90°$ ACW. Let me recheck the order.
First: reflection in x-axis ($M_1$). Second: reflection in $y=x$ ($M_2$).
This is $90°$ anti-clockwise. So the correct answer is (a) Rotation of $90°$ ACW.
Correction: Always be careful with matrix multiplication order!
Q5. A triangle has area $20$ cm². After an enlargement with scale factor $-\dfrac{1}{2}$, the area of the image is:
(a) $-10$ cm² (b) $5$ cm² (c) $10$ cm² (d) $40$ cm²
Answer: (b) $5$ cm²
Area is always positive! The negative sign in $k$ affects orientation, not area.
Fill in the Blank
Q6. The image of the point $(a, b)$ under reflection in the origin is ____.
Answer: $(-a, -b)$
Q7. Two successive reflections in parallel lines that are $5$ units apart is equivalent to a translation of ____ units.
Answer: $10$ units
Translation distance $= 2 \times 5 = 10$ units, in the direction perpendicular to the lines (from first mirror towards second mirror).
Q8. The determinant of any reflection matrix is ____.
Answer: $-1$
All reflections have $\det = -1$, indicating area is preserved but orientation is reversed.
Q9. Under a rotation of $270°$ anti-clockwise about the origin, the point $(3, 0)$ maps to ____.
Answer: $(0, -3)$
$270°$ ACW $= 90°$ CW: $(x,y) \to (y, -x) = (0, -3)$
Q10. The inverse of the enlargement matrix $\begin{pmatrix}4&0\\0&4\end{pmatrix}$ is ____.
Answer: $\begin{pmatrix}\frac{1}{4}&0\\0&\frac{1}{4}\end{pmatrix}$
The inverse of enlargement with scale factor $k$ is enlargement with scale factor $\frac{1}{k}$.
Short Answer Questions
Q11. Find the image of the circle $(x-2)^2 + (y-3)^2 = 25$ under a translation by $\begin{pmatrix}-1\\4\end{pmatrix}$.
Answer:
Replace $x$ with $x+1$ and $y$ with $y-4$:
The circle moves: centre from $(2,3)$ to $(1,7)$, radius stays $5$. ✓
Q12. $A(1, 3)$ maps to $A'(-3, 1)$ and $B(4, 0)$ maps to $B'(0, 4)$ under a single transformation. Describe the transformation fully.
Answer:
Notice the pattern: $(1,3) \to (-3,1)$ and $(4,0) \to (0,4)$. It looks like $(x,y) \to (-y, x)$, which is rotation of $90°$ anti-clockwise.
Verification:
Q13. Find the image of $(2, -1)$ under the transformation with matrix $\begin{pmatrix}-3&0\\0&-3\end{pmatrix}$ and describe the transformation.
Answer:
Image: $(-6, 3)$. This is an enlargement with scale factor $-3$ about the origin. The negative scale factor means the image is on the opposite side of the origin and three times as far.
Q14. The line $y = 3x – 1$ is reflected in the line $y = -x$. Find the equation of the image line.
Answer:
Reflection in $y = -x$: swap $x$ and $y$ AND change both signs. So replace $x$ with $-y$ and $y$ with $-x$:
Step-by-Step Calculation Questions
Q15. Find the single $2 \times 2$ matrix representing a reflection in the x-axis followed by a rotation of $90°$ anti-clockwise about the origin. Verify by applying this matrix to the point $(2, 3)$ using both the step-by-step method and the composite matrix.
Answer:
Step-by-step:
Reflection in x-axis: $(2, 3) \to (2, -3)$
Rotation $90°$ ACW: $(2, -3) \to (3, 2)$
Composite matrix:
Note: The composite matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is the reflection in $y = x$ matrix! So reflecting in the x-axis then rotating $90°$ ACW gives the same result as reflecting in $y = x$.
Q16. Find the image of the line $2x – 3y + 6 = 0$ under the transformation represented by the matrix $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Interpret the result geometrically.
Answer:
The matrix gives $x’ = x$ and $y’ = -y$, so $x = x’$ and $y = -y’$. Substitute:
Geometric interpretation: The matrix $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ represents reflection in the x-axis. The original line $2x – 3y + 6 = 0$ has slope $\frac{2}{3}$ and y-intercept $2$. The image $2x + 3y + 6 = 0$ has slope $-\frac{2}{3}$ and y-intercept $-2$. The slopes are negative reciprocals of each other? No — the slope changed from $\frac{2}{3}$ to $-\frac{2}{3}$, which makes sense because reflection in the x-axis negates the y-values, changing the sign of the slope.
Q17. Triangle $PQR$ has vertices $P(2, 1)$, $Q(6, 1)$, $R(4, 5)$. Find the image under enlargement with scale factor $\dfrac{1}{2}$ about the point $(2, 1)$. Hence find the ratio of the area of the image to the area of the original.
Answer:
Centre $C(2, 1)$, scale factor $k = \dfrac{1}{2}$:
Original area: base $PQ = 4$, height from $R$ to $PQ$ $= 4$. Area $= \frac{1}{2} \times 4 \times 4 = 8$.
Image area: base $P’Q’ = 2$, height from $R’$ to $P’Q’$ $= 2$. Area $= \frac{1}{2} \times 2 \times 2 = 2$.
Q18. The transformation $T$ is defined by the matrix $M = \begin{pmatrix}0&1\\1&0\end{pmatrix}$. (a) Identify $T$. (b) Find $T^{-1}$. (c) Find the image of the triangle with vertices $(1, 2)$, $(3, 4)$, $(5, 0)$ under $T$. (d) Find the equation of the image of the line $y = 2x + 1$ under $T$.
Answer:
(a) $M = \begin{pmatrix}0&1\\1&0\end{pmatrix}$ is reflection in the line $y = x$. ✓
(b) Since reflection is self-inverse: $M^{-1} = M = \begin{pmatrix}0&1\\1&0\end{pmatrix}$. Verification: $M^2 = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I$ ✓
(c) Swap each pair of coordinates: $(1,2) \to (2,1)$, $(3,4) \to (4,3)$, $(5,0) \to (0,5)$
(d) Swap $x$ and $y$ in $y = 2x + 1$: $x = 2y + 1 \Rightarrow 2y = x – 1 \Rightarrow y = \dfrac{x-1}{2}$
Q19. Find the single matrix representing a rotation of $90°$ anti-clockwise about the origin followed by a reflection in the y-axis. Hence find the image of $(3, -2)$.
Answer:
Image: $(-2, 3)$. Interestingly, this composite equals reflection in $y = x$!
Q20. Prove that reflecting in the line $y = x$ followed by reflecting in the line $y = -x$ is equivalent to a rotation of $180°$ about the origin.
Answer:
This is the matrix for rotation of $180°$ about the origin (or reflection in the origin). ✓
Alternative proof using geometry: The lines $y = x$ and $y = -x$ intersect at the origin at an angle of $90°$. Two reflections in intersecting lines with angle $\alpha$ between them equals a rotation of $2\alpha = 2 \times 90° = 180°$ about the intersection point. ✓
Q21. A transformation maps the unit square with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$ to a parallelogram with vertices $(0,0)$, $(2,1)$, $(1,3)$, $(-1,2)$. (a) Find the transformation matrix. (b) Find the determinant and interpret. (c) Find the area of the image.
Answer:
(a) The images of $(1,0)$ and $(0,1)$ give the columns of the matrix:
Verification: $M\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}2-1\\1+3\end{pmatrix} = \begin{pmatrix}1\\4\end{pmatrix}$. But the given image of $(1,1)$ is $(1,3)$, not $(1,4)$. This means this is NOT a linear transformation! Let me re-examine.
Actually, checking all vertices: $(0,0) \to (0,0)$ ✓. If this is linear, then $(1,1) = (1,0)+(0,1)$ should map to $(2,1)+(-1,3) = (1,4)$. But the given image is $(1,3)$. There is an inconsistency, which means either the transformation is not linear or the data has an error.
If we use $(1,1) \to (1,3)$ as additional info, we need a translation component, suggesting this is an affine transformation (linear + translation). This would require the $3 \times 3$ homogeneous matrix approach.
This question illustrates the importance of checking consistency when finding transformation matrices!
Q22. Point $P(3, 2)$ is first translated by $\begin{pmatrix}-2\\5\end{pmatrix}$, then reflected in the y-axis, then rotated $90°$ anti-clockwise about the origin. Find the final image of $P$.
Answer (step by step):
Step 1 — Translation: $(3, 2) \to (3-2, 2+5) = (1, 7)$
Step 2 — Reflection in y-axis: $(1, 7) \to (-1, 7)$
Step 3 — Rotation $90°$ ACW: $(-1, 7) \to (-7, -1)$
Note: Since translation is not a $2 \times 2$ matrix operation, we cannot combine all three into a single $2 \times 2$ matrix. We would need $3 \times 3$ homogeneous matrices for that.
Q23. Find the image of the line $x – y + 4 = 0$ under a rotation of $180°$ about the point $(-1, 2)$.
Answer:
Method: Find two points on the line, rotate each about $(-1, 2)$ by $180°$, then find the image line.
Point 1: when $x = 0$: $y = 4$, so $(0, 4)$. Rotate about $(-1, 2)$ by $180°$:
Point 2: when $y = 0$: $x = -4$, so $(-4, 0)$. Rotate about $(-1, 2)$ by $180°$:
Line through $(-2, 0)$ and $(2, 4)$: slope $= \dfrac{4-0}{2-(-2)} = 1$
Q24. The transformation $T_1$ is “enlargement with scale factor $2$ about the origin” and $T_2$ is “rotation of $90°$ anti-clockwise about the origin.” Does $T_1 \circ T_2 = T_2 \circ T_1$? Prove your answer.
Answer:
Both give the same matrix! So YES, $T_1 \circ T_2 = T_2 \circ T_1$ in this case.
Why? Because enlargement uses a scalar matrix ($kI$), and scalar matrices commute with ALL matrices: $(kI)M = M(kI) = kM$. This is a special property of scalar matrices.
Q25. Find the equation of the mirror line if the point $(3, 7)$ is reflected to $(11, -1)$.
Answer:
The mirror line is $y = x – 4$.