A transformation moves or resizes a shape. The original is the object; the result is the image.
| Transformation | What it does | You must state | Congruent? |
|---|---|---|---|
| Translation | Slides the shape | The column vector | Yes |
| Reflection | Flips it in a mirror line | The equation of the mirror line | Yes |
| Rotation | Turns it about a point | Angle, direction, centre | Yes |
| Enlargement | Resizes it | Scale factor and centre | No (unless SF is $\pm 1$) |
A translation slides every point the same distance in the same direction. It is described by a column vector.
Translate the point $(4,\ -1)$ by the vector $\begin{pmatrix} -3 \\ 5 \end{pmatrix}$.
Image: $(1,\ 4)$
Triangle $P$ has a vertex at $(2,\ 7)$. After a translation, the matching vertex of triangle $Q$ is at $(-4,\ 3)$. Find the translation vector.
Vector: $\begin{pmatrix} -6 \\ -4 \end{pmatrix}$
A reflection flips a shape across a mirror line. Every point and its image are the same distance from the mirror, on opposite sides, measured perpendicular to the line.
| Mirror line | Effect on $(x,\ y)$ |
|---|---|
| The $x$-axis, $y = 0$ | $(x,\ -y)$ — the $y$-coordinate changes sign |
| The $y$-axis, $x = 0$ | $(-x,\ y)$ — the $x$-coordinate changes sign |
| $x = a$ (a vertical line) | $(2a - x,\ y)$ |
| $y = b$ (a horizontal line) | $(x,\ 2b - y)$ |
| $y = x$ | $(y,\ x)$ — swap the coordinates |
| $y = -x$ | $(-y,\ -x)$ — swap and change both signs |
- Draw the mirror line accurately (a dashed line).
- For each vertex, count the perpendicular distance to the mirror.
- Count the same distance beyond the mirror and mark the image point.
- Join the image points in the same order.
Reflect the triangle with vertices $(1,\ 4)$, $(3,\ 4)$ and $(1,\ 7)$ in the line $y = x$.
A shape is reflected so that $(2,\ 5)$ maps to $(2,\ -3)$ and $(6,\ 5)$ maps to $(6,\ -3)$. Find the mirror line.
Mirror line: $y = 1$
A rotation turns a shape about a fixed point called the centre of rotation.
| Rotation about the origin | Effect on $(x,\ y)$ |
|---|---|
| $90^\circ$ anticlockwise | $(-y,\ x)$ |
| $90^\circ$ clockwise | $(y,\ -x)$ |
| $180^\circ$ | $(-x,\ -y)$ |
- Mark the centre of rotation clearly.
- Use tracing paper: trace the shape and the centre, hold a pencil on the centre, and turn.
- If no tracing paper, count the "across and up" from the centre to a vertex, then rotate that pair of steps.
- Mark all the image vertices and join them.
Rotate the point $(5,\ 3)$ by $90^\circ$ clockwise about the centre $(1,\ 1)$.
Check with the formula: relative coordinates $(4, 2)$; clockwise $90^\circ$ gives $(y, -x) = (2, -4)$; add the centre: $(1+2,\ 1-4) = (3, -3)$ ✓
Describe how to find the centre when you can see the object and its image.
This works because the centre is equidistant from every point and its image.
An enlargement changes the size of a shape. The image is similar to the object, not congruent.
| Scale factor $k$ | Effect |
|---|---|
| $k > 1$ | Bigger, same way up, same side of the centre |
| $0 \lt k \lt 1$ | Smaller (still called an enlargement), same way up |
| $k = -1$ | Same size, upside down — equivalent to a $180^\circ$ rotation about the centre |
| $k \lt 0$ | Upside down and on the opposite side of the centre |
- Draw a ray from the centre through each vertex of the object.
- Measure the distance from the centre to each vertex.
- Multiply each distance by the scale factor.
- Mark the new points along the rays (backwards through the centre if $k$ is negative).
- Join the image vertices.
Enlarge the point $(4,\ 3)$ by scale factor $3$, centre $(1,\ 1)$.
Enlarge the point $(5,\ 4)$ by scale factor $-2$, centre $(1,\ 2)$.
The image is twice the size, upside down, and on the opposite side of the centre.
A triangle with sides $3$ cm, $4$ cm and $5$ cm is enlarged to one with sides $7.5$ cm, $10$ cm and $12.5$ cm. Find the scale factor, and describe how to locate the centre.
Translation
State the column vector. Image $=$ object $+$ vector.
Reflection
State the equation of the mirror line, e.g. $y = -2$.
Rotation
State angle, direction and centre. $180^\circ$ needs no direction.
Enlargement
State scale factor and centre.
$y = x$
Reflection swaps the coordinates: $(x,y) \to (y,x)$.
$90^\circ$ about O
Anticlockwise $(x,y)\to(-y,x)$; clockwise $(x,y)\to(y,-x)$.
Negative SF
Image is inverted and on the opposite side of the centre.
Congruence
Translation, reflection and rotation preserve size; enlargement does not.
Finding a centre
Join matching vertices; for enlargement the lines meet at the centre.
Translate the point $(-2,\ 6)$ by the vector $\begin{pmatrix} 5 \\ -8 \end{pmatrix}$.
▶ Show solution
$x$: $-2 + 5 = 3$
$y$: $6 + (-8) = -2$
Image: $(3,\ -2)$
Reflect the point $(3,\ -5)$ in (a) the $x$-axis, (b) the $y$-axis, (c) the line $y = x$.
▶ Show solution
(a) $x$-axis: change the sign of $y$ → $(3,\ 5)$
(b) $y$-axis: change the sign of $x$ → $(-3,\ -5)$
(c) $y = x$: swap the coordinates → $(-5,\ 3)$
Rotate the point $(4,\ 2)$ by $180^\circ$ about the origin, then by $90^\circ$ anticlockwise about the origin (starting again from $(4, 2)$).
▶ Show solution
$180^\circ$: $(x,y) \to (-x,-y)$, giving $(-4,\ -2)$.
$90^\circ$ anticlockwise: $(x,y) \to (-y,x)$, giving $(-2,\ 4)$.
Describe fully the single transformation that maps $(1,\ 3)$ to $(1,\ -1)$ and $(4,\ 5)$ to $(4,\ -3)$.
▶ Show solution
The $x$-coordinates are unchanged, so the mirror line is horizontal.
Midpoint of the first pair of $y$-values: $\dfrac{3 + (-1)}{2} = 1$.
Check with the second pair: $\dfrac{5 + (-3)}{2} = 1$ ✓ — the same line.
A reflection in the line $y = 1$.
Why not a translation? The first point moves down $4$ but the second moves down $8$, so no single translation works.
Enlarge the point $(6,\ 8)$ by scale factor $\tfrac{1}{2}$, centre the origin.
▶ Show solution
With the centre at the origin, simply multiply both coordinates by the scale factor.
$(6 \times \tfrac{1}{2},\ 8 \times \tfrac{1}{2}) = (3,\ 4)$
Enlarge the point $(7,\ 5)$ by scale factor $2$, centre $(3,\ 1)$.
▶ Show solution
Step from centre: $7 - 3 = 4$ across, $5 - 1 = 4$ up.
Multiply by $2$: $8$ across, $8$ up.
Image $= (3 + 8,\ 1 + 8) = (11,\ 9)$
Enlarge the point $(4,\ 6)$ by scale factor $-3$, centre $(1,\ 2)$.
▶ Show solution
Step from centre: $3$ across, $4$ up.
Multiply by $-3$: $-9$ across, $-12$ up.
Image $= (1 - 9,\ 2 - 12) = (-8,\ -10)$
The image is $3$ times the size, inverted, and on the opposite side of the centre.
A shape is enlarged by scale factor $4$. Its original area was $7\text{ cm}^2$. Find the area of the image.
▶ Show solution
Lengths are multiplied by $4$, so areas are multiplied by $4^2 = 16$.
Image area $= 7 \times 16 = 112\text{ cm}^2$
Triangle $T$ has vertices $(2,\ 1)$, $(2,\ 4)$ and $(4,\ 1)$. Triangle $U$ has vertices $(-1,\ 2)$, $(-4,\ 2)$ and $(-1,\ 4)$. Describe fully the single transformation that maps $T$ onto $U$.
▶ Show solution
Check the side lengths: $T$ has legs $3$ and $2$; $U$ has legs $3$ and $2$. Same size, so it is not an enlargement.
Test a $90^\circ$ anticlockwise rotation about the origin, $(x,y) \to (-y,x)$:
$(2,1) \to (-1,2)$ ✓ $(2,4) \to (-4,2)$ ✓ $(4,1) \to (-1,4)$ ✓
All three vertices match.
A rotation of $90^\circ$ anticlockwise about the origin $(0,\ 0)$.
Shape $A$ has vertices $(2,\ 2)$, $(4,\ 2)$ and $(2,\ 5)$. Shape $B$ has vertices $(8,\ 2)$, $(12,\ 2)$ and $(8,\ 8)$.
(a) Show that $B$ is an enlargement of $A$ and find the scale factor. (b) Find the centre of enlargement. (c) Find the ratio of the areas.
▶ Show solution
(a) $A$ has a horizontal side of $2$ and a vertical side of $3$. $B$ has $4$ and $6$.
$\dfrac{4}{2} = 2$ and $\dfrac{6}{3} = 2$ — both give the same factor, so it is an enlargement with scale factor $2$.
(b) Let the centre be $(a,\ b)$. For scale factor $2$: image $=$ centre $+ 2 \times$ (object $-$ centre).
Using $(2,2) \to (8,2)$ for the $x$-coordinate: $8 = a + 2(2 - a) = 4 - a$, so $a = -4$.
For $y$: $2 = b + 2(2 - b) = 4 - b$, so $b = 2$.
Centre $= (-4,\ 2)$.
Check with $(2,5)\to(8,8)$: $y$: $2 + 2(5-2) = 8$ ✓
(c) Areas scale by $k^2 = 4$, so the ratio of areas $A : B = 1 : 4$.