Chapter 6: Transformations
A transformation moves or changes a shape according to a precise rule. The original shape is called the object and the result is called the image. The four transformations studied at IGCSE are translation, rotation, reflection, and enlargement. In every exam question you must both perform transformations and describe them fully — each type requires specific information.
Exam Board Coverage
All four transformations appear at both Foundation and Extended tier. Combined transformations, negative enlargement, and fractional scale factors are primarily Extended. The key rule: use one word only to name the transformation (Translation, Rotation, Reflection, or Enlargement) — never use alternatives like "moved" or "flipped".
6.1 Translation
In a translation, every point of the shape moves by the same displacement. The shape does not rotate, reflect, or resize. Every point shifts by the same horizontal amount and the same vertical amount.
Column Vector Notation
A translation is described by a column vector $\dbinom{a}{b}$, where $a$ is the shift in the $x$-direction (positive = right, negative = left) and $b$ is the shift in the $y$-direction (positive = up, negative = down).
If point $P(x, y)$ is translated by $\dbinom{a}{b}$, its image is $P'(x+a,\; y+b)$.
Example 6.1 — Translating a shape
Triangle $A$ has vertices at $(1,1)$, $(3,1)$, $(2,3)$. Translate it by $\dbinom{3}{-2}$.
Step 1 Add $3$ to every $x$-coordinate and $-2$ to every $y$-coordinate.
Step 2 $(1,1) \to (4,-1)$, $(3,1) \to (6,-1)$, $(2,3) \to (5,1)$
Image triangle $A'$ has vertices at $(4,-1)$, $(6,-1)$, $(5,1)$.
Example 6.2 — Finding the translation vector
Shape $P$ has a vertex at $(5, 4)$. Its image $P'$ has the corresponding vertex at $(2, 7)$. Find the translation vector.
$\dbinom{a}{b} = \dbinom{2-5}{7-4} = \dbinom{-3}{3}$
Check: $5 + (-3) = 2$ ✓ and $4 + 3 = 7$ ✓.
6.2 Rotation
A rotation turns a shape through a given angle about a fixed point called the centre of rotation. Every point moves along an arc of a circle centred at that point. The shape and size are preserved.
Describing a Rotation
To describe a rotation fully you need three pieces of information:
- The word "Rotation"
- The angle (e.g., 90°, 180°, 270°)
- The direction (clockwise or anticlockwise) — not needed for 180°
- The centre of rotation (coordinates)
Rotating Points
For a rotation of $90°$ anticlockwise about the origin: $(x, y) \to (-y, x)$.
For a rotation of $90°$ clockwise about the origin: $(x, y) \to (y, -x)$.
For a rotation of $180°$ about the origin: $(x, y) \to (-x, -y)$.
Example 6.3 — Rotating 90° anticlockwise about the origin
Rotate point $A(4, 1)$ by $90°$ anticlockwise about the origin.
$(x, y) \to (-y, x)$: $(4, 1) \to (-1, 4)$. Image: $A'(-1, 4)$.
Example 6.4 — Finding the centre of rotation
Shape $T$ is mapped to $T'$ by a rotation. To find the centre: draw the perpendicular bisectors of the segments joining each vertex to its image. The perpendicular bisectors all pass through the centre of rotation.
6.3 Reflection
A reflection maps each point to its mirror image across a mirror line. Every point and its image are equidistant from the mirror line, and the line joining them is perpendicular to the mirror line.
Describing a Reflection
Two pieces of information needed:
- The word "Reflection"
- The equation of the mirror line
Common Mirror Lines
- $x$-axis ($y = 0$): $(x, y) \to (x, -y)$
- $y$-axis ($x = 0$): $(x, y) \to (-x, y)$
- $y = x$: $(x, y) \to (y, x)$
- $y = -x$: $(x, y) \to (-y, -x)$
- $x = a$ (vertical): $(x, y) \to (2a - x,\; y)$
- $y = b$ (horizontal): $(x, y) \to (x,\; 2b - y)$
Example 6.5 — Reflecting in $y = x$
Reflect point $P(3, -1)$ in the line $y = x$.
$(x, y) \to (y, x)$: $(3, -1) \to (-1, 3)$. Image: $P'(-1, 3)$.
Example 6.6 — Finding the mirror line
Point $A(2, 5)$ maps to $A'(6, 5)$. The $y$-coordinates are equal, so the mirror line is vertical. The midpoint of $AA'$ is $(4, 5)$, so the mirror line is $x = 4$.
6.4 Enlargement (Positive Scale Factor)
An enlargement changes the size of a shape while keeping all angles the same. Every distance from the centre of enlargement is multiplied by the scale factor $k$.
Enlargement with Positive Scale Factor
- $k > 1$: the image is larger than the object.
- $0 < k < 1$: the image is smaller (a reduction).
- $k = 1$: the image is identical to the object (no change).
If the centre of enlargement is $C(c_x, c_y)$ and scale factor is $k$, then for any object point $P(x,y)$:
$$P' = C + k(P - C) = \bigl(c_x + k(x-c_x),\; c_y + k(y-c_y)\bigr)$$
Example 6.7 — Enlargement with centre not at origin
Enlarge triangle with vertices $A(2,1)$, $B(4,1)$, $C(3,3)$ by scale factor $2$ with centre of enlargement $O(0,0)$.
Multiply each coordinate by $2$: $A'(4,2)$, $B'(8,2)$, $C'(6,6)$.
6.5 Enlargement (Negative and Fractional Scale Factor) Extended
Fractional Scale Factor
If $0 < k < 1$, the image is on the same side as the object relative to the centre, but smaller. All angles are preserved and the shape is similar to the original.
Negative Scale Factor
If $k < 0$, the image is inverted and on the opposite side of the centre from the object. The rays from the centre go through the centre and continue to the image on the other side.
Example 6.8 — Negative scale factor
Point $P(4, 2)$, centre of enlargement $C(1, 1)$, scale factor $-2$.
Step 1 Vector from centre to $P$: $(4-1, 2-1) = (3, 1)$.
Step 2 Multiply by $-2$: $(-6, -2)$.
Step 3 Add to centre: $P' = (1-6, 1-2) = (-5, -1)$.
6.6 Describing a Transformation Fully
Required Information for Each Transformation
| Transformation | Must State |
|---|---|
| Translation | The word "Translation" + column vector $\binom{a}{b}$ |
| Rotation | "Rotation" + angle + direction (CW/ACW) + centre coordinates |
| Reflection | "Reflection" + equation of mirror line |
| Enlargement | "Enlargement" + scale factor + centre coordinates |
Example 6.9 — Describing a transformation
Shape $A$ has vertices $(1,1)$, $(3,1)$, $(3,2)$. Shape $A'$ has vertices $(-1,-1)$, $(-3,-1)$, $(-3,-2)$. Describe fully the single transformation.
Step 1 Check: all coordinates are negated. This is a rotation of $180°$ about the origin.
Describe Rotation, $180°$, about the origin $(0, 0)$. (No direction needed for $180°$.)
6.7 Combined Transformations Extended
When two transformations are applied in sequence, the order matters. Transformation $T_2$ followed by $T_1$ generally gives a different result from $T_1$ followed by $T_2$. The combined effect can often be described as a single transformation.
Example 6.10 — Two reflections
Shape $A$ is reflected in the $x$-axis to give $A'$, then $A'$ is reflected in the $y$-axis to give $A''$. Describe the single transformation mapping $A$ to $A''$.
Step 1 Reflection in $x$-axis: $(x,y) \to (x,-y)$.
Step 2 Reflection in $y$-axis: $(x,-y) \to (-x,-y)$.
Result $(x,y) \to (-x,-y)$ — this is a rotation of $180°$ about the origin.
6.8 Congruence and Similarity
Congruence
Two shapes are congruent if they are identical in shape and size. Transformations that preserve congruence: translation, rotation, and reflection. The image always has the same area and the same side lengths as the object.
Similarity
Two shapes are similar if they have the same shape (all angles equal) but different sizes. Enlargement (any scale factor $\neq 1$) produces a similar (not congruent) image.
- Corresponding lengths are in the ratio $k : 1$.
- Corresponding areas are in the ratio $k^2 : 1$.
- Corresponding volumes are in the ratio $k^3 : 1$.
Exercises
Exercise 6.1
Triangle $T$ has vertices $A(1,2)$, $B(3,2)$, $C(2,4)$. Translate $T$ by $\dbinom{-4}{1}$. Write the coordinates of the image.
Show Solution
$A(1,2) \to A'(-3,3)$; $B(3,2) \to B'(-1,3)$; $C(2,4) \to C'(-2,5)$
Exercise 6.2
Point $P(5,3)$ maps to $P'(1,6)$ under a translation. Find the translation vector.
Show Solution
$\dbinom{1-5}{6-3} = \dbinom{-4}{3}$
Exercise 6.3
Rotate point $A(3, 5)$ by $90°$ clockwise about the origin. Write the image coordinates.
Show Solution
$90°$ clockwise: $(x,y) \to (y,-x)$. $A(3,5) \to A'(5,-3)$.
Exercise 6.4
Reflect point $Q(4,-2)$ in the line $y = -x$.
Show Solution
Reflection in $y = -x$: $(x,y) \to (-y,-x)$. $Q(4,-2) \to Q'(2,-4)$.
Exercise 6.5
Enlarge shape with vertex $P(3,2)$ by scale factor $3$ with centre of enlargement $(1,1)$. Find the image of $P$.
Show Solution
$P' = (1 + 3(3-1),\; 1 + 3(2-1)) = (1+6,\; 1+3) = (7, 4)$
Exercise 6.6
Shape $A$ has vertices $(2,1)$, $(4,1)$, $(4,3)$. Shape $A'$ has vertices $(-2,-1)$, $(-4,-1)$, $(-4,-3)$. Describe fully the single transformation.
Show Solution
All coordinates are negated: $(x,y) \to (-x,-y)$.
Rotation, 180°, about the origin (0, 0).
Exercise 6.7
Describe fully the single transformation that maps shape $S$ with vertices $(1,1)$, $(2,1)$, $(2,3)$ to shape $S'$ with vertices $(1,-1)$, $(2,-1)$, $(2,-3)$.
Show Solution
$y$-coordinates are negated: $(x,y) \to (x,-y)$.
Reflection in the $x$-axis (line $y=0$).
Exercise 6.8 Extended
Shape $R$ has vertices $(2,0)$, $(4,0)$, $(4,2)$. Enlarge $R$ by scale factor $-\tfrac{1}{2}$ with centre $(0,0)$. Write the image vertices.
Show Solution
Multiply each coordinate by $-\tfrac{1}{2}$: $(2,0)\to(-1,0)$; $(4,0)\to(-2,0)$; $(4,2)\to(-2,-1)$.
Exercise 6.9 Extended
Shape $A$ is reflected in the line $x=2$ to give $A'$, then $A'$ is reflected in the line $x=5$ to give $A''$. Describe the single transformation mapping $A$ to $A''$.
Show Solution
Two reflections in parallel vertical lines: distance between lines $= 5-2 = 3$. The combined effect is a translation of $\dbinom{2 \times 3}{0} = \dbinom{6}{0}$ (i.e., 6 units to the right).
Exercise 6.10
Two similar shapes have corresponding side lengths $6\,\text{cm}$ and $9\,\text{cm}$. The area of the smaller shape is $20\,\text{cm}^2$. Find the area of the larger shape.
Show Solution
Scale factor $k = 9/6 = 1.5$. Area ratio $= k^2 = 2.25$.
Larger area $= 20 \times 2.25 = \mathbf{45\,\text{cm}^2}$.
Exam Tips
Tip 1 — Use the correct transformation word
The transformation type must be named with exactly the right word: Translation, Rotation, Reflection, or Enlargement. Never say "moved", "shifted", "flipped", or "turned" — you will lose the mark for naming the type.
Tip 2 — Rotation: state all four elements
For a rotation: the word, the angle, the direction (CW or ACW — not required for 180°), and the coordinates of the centre. Missing any one of these loses a mark.
Tip 3 — Enlargement scale factor can be calculated
To find the scale factor of an enlargement, divide any image length by the corresponding object length. To find the centre, draw lines through each object point and its image — they all meet at the centre.
Tip 4 — Negative scale factor: image is inverted
A negative scale factor rotates the image through $180°$ as well as scaling it. The image appears "upside down" relative to the object, on the opposite side of the centre of enlargement.