Basic Transformations Revision Notes | GCSE AQA Higher Maths

1. Why Transformations Matter

Every time a designer tiles a floor, a programmer animates a character, or an architect mirrors a building plan, they are using transformations. In GCSE Maths, transformations are the rules that move, flip, turn, or resize shapes on a coordinate grid — and examiners test whether you can both perform them and describe them precisely.

There are four transformations you need to master: translation, reflection, rotation, and enlargement. Each one has its own set of information you must state to describe it fully — miss one piece and you lose marks.

2. Translation

A translation slides every point of a shape by exactly the same amount. Because nothing rotates or flips, the image is congruent to the object — identical in size and shape, just in a different position.

To describe a translation fully, you need one thing only: the column vector. For example, $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ means move 3 right and 2 down.

Worked example: Performing a Translation

Triangle $A$ has vertices at $(1, 1)$, $(3, 1)$, and $(2, 3)$. Translate it by $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$.

1. $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$: move 4 left, 2 up — Identify the horizontal and vertical components before touching any coordinates. [M1]
2. $(1 + (-4),\; 1 + 2) = (-3, 3)$ — Add the vector components to the $x$ and $y$ coordinates separately.
3. $(3 + (-4),\; 1 + 2) = (-1, 3)$ — Same vector applied to every vertex — this is what makes it a translation.
4. $(2 + (-4),\; 3 + 2) = (-2, 5)$ — All three vertices shift by the same amount. [A1]
5. Image vertices: $(-3, 3)$, $(-1, 3)$, $(-2, 5)$ — Label the image $A'$ to distinguish it from the object. The shape is congruent to the original.

Answer: Image vertices at $(-3, 3)$, $(-1, 3)$, and $(-2, 5)$

3. Reflection

A reflection flips a shape across a mirror line. Each point on the object maps to a point on the image that is the same perpendicular distance from the mirror line, but on the opposite side. The image is congruent to the object, but its orientation is reversed — like a left hand becoming a right hand.

To describe a reflection fully, you must state the equation of the mirror line. Common mirror lines at GCSE include $y = 0$ (the $x$-axis), $x = 0$ (the $y$-axis), $y = x$, $y = -x$, and vertical or horizontal lines such as $x = 3$.

Worked example: Reflecting a Shape in the Line $y = x$

Point $P$ is at $(2, 5)$. Find its image after reflection in the line $y = x$.

1. Reflecting in $y = x$: swap the $x$ and $y$ coordinates — This rule works because the line $y = x$ is the set of all points where $x = y$, so swapping coordinates gives the mirror image. [M1]
2. $P(2, 5) \rightarrow P'(5, 2)$ — Swap: $x = 2$ becomes the new $y$, and $y = 5$ becomes the new $x$. [A1]
3. Midpoint of $PP'$: $\left(\frac{2+5}{2}, \frac{5+2}{2}\right) = (3.5, 3.5)$, which lies on $y = x$ ✓ — A sanity check: the midpoint of any point and its image must lie on the mirror line.

Answer: $P' = (5, 2)$

4. Rotation

A rotation turns a shape through an angle about a fixed point called the centre of rotation. The image is congruent to the object — lengths and angles are preserved — but the orientation changes (unless the angle is $360^\circ$).

To describe a rotation fully, you need three things: the angle of rotation, the direction (clockwise or anticlockwise), and the coordinates of the centre of rotation. Miss any one of these and the description is incomplete.

Worked example: Rotating a Shape $90^\circ$ Clockwise About the Origin

Triangle $B$ has vertices at $(1, 2)$, $(4, 2)$, and $(4, 5)$. Rotate it $90^\circ$ clockwise about the origin.

1. $90^\circ$ clockwise: $(x, y) \rightarrow (y, -x)$ — Memorise this rule or derive it by thinking about where the positive $x$-axis maps to — it rotates to point downwards, becoming the negative $y$-direction. [M1]
2. $(1, 2) \rightarrow (2, -1)$ — New $x = $ old $y = 2$; new $y = -$old $x = -1$.
3. $(4, 2) \rightarrow (2, -4)$ — New $x = 2$; new $y = -4$.
4. $(4, 5) \rightarrow (5, -4)$ — New $x = 5$; new $y = -4$. [A1]
5. Image vertices: $(2, -1)$, $(2, -4)$, $(5, -4)$ — The image sits in the fourth quadrant. Check: the shape is congruent — side lengths are unchanged.

Answer: Image vertices at $(2, -1)$, $(2, -4)$, and $(5, -4)$

5. Enlargement

An enlargement scales a shape by a scale factor from a fixed centre of enlargement. Unlike the other three transformations, enlargement can change the size of the shape. When |k| = 1, the image is congruent to the object; when |k| ≠ 1, the image is similar to the object — same angles, proportional sides.

To describe an enlargement fully, you need two things: the scale factor and the coordinates of the centre of enlargement. A scale factor greater than 1 makes the image larger; a scale factor between 0 and 1 makes it smaller; a negative scale factor flips the image through the centre.

Worked example: Performing an Enlargement with a Given Centre

Square $C$ has vertices at $(1, 1)$, $(3, 1)$, $(3, 3)$, and $(1, 3)$. Enlarge it by scale factor $2$ from centre $(0, 0)$.

1. Multiply each coordinate by the scale factor when the centre is the origin — When the centre of enlargement is the origin, image coordinates $= k \times$ object coordinates. [M1]
2. $2 \times (1, 1) = (2, 2)$ — Both $x$ and $y$ are multiplied by 2.
3. $2 \times (3, 1) = (6, 2)$ — $x: 2 \times 3 = 6$; $y: 2 \times 1 = 2$.
4. $2 \times (3, 3) = (6, 6)$ — $x: 2 \times 3 = 6$; $y: 2 \times 3 = 6$.
5. $2 \times (1, 3) = (2, 6)$ — Image vertices: $(2,2)$, $(6,2)$, $(6,6)$, $(2,6)$. [A1]
6. Object side: $3 - 1 = 2$ units. Image side: $6 - 2 = 4$ units. $4 = 2 \times 2$ ✓ — Each side of the image is exactly twice the corresponding side of the object, confirming scale factor 2.

Answer: Image vertices at $(2, 2)$, $(6, 2)$, $(6, 6)$, and $(2, 6)$

When the centre of enlargement is not at the origin, use a ray method: draw a line from the centre through each vertex of the object, then measure along that ray. The image vertex is at a distance of $k \times$ (distance from centre to object vertex) from the centre.

Worked example: Enlargement from a Non-Origin Centre

Point $P$ is at $(5, 4)$. Enlarge it by scale factor $3$ from centre $C = (2, 1)$.

1. Vector $\overrightarrow{CP} = (5 - 2,\; 4 - 1) = (3, 3)$ — Subtract the centre's coordinates from the object's coordinates. [M1]
2. $3 \times (3, 3) = (9, 9)$ — Scale the displacement vector by $k = 3$.
3. Image $P' = (2 + 9,\; 1 + 9) = (11, 10)$ — Adding back to the centre gives the image position. [A1]
4. $|CP| = \sqrt{3^2 + 3^2} = \sqrt{18}$; $|CP'| = \sqrt{9^2 + 9^2} = \sqrt{162} = 3\sqrt{18}$ ✓ — The distance from centre to image is exactly 3 times the distance from centre to object.

Answer: $P' = (11, 10)$

6. Congruence, Similarity, and Describing Transformations

Translations, reflections, and rotations all produce images that are congruent to the original — same size, same shape, same angles. Enlargements (with scale factor $ eq \pm 1$) produce images that are similar — same angles, but proportionally different side lengths.

When an exam question asks you to describe a transformation, you must identify the type first, then give all the required information. A partial description — for example, saying "rotation" without the angle, direction, and centre — will not earn full marks.

7. Practice and Exam Technique

Examiners award marks for precision. The worked examples above show the method — now it is time to apply it independently before looking at the most common ways marks are dropped.

Additional core detail (the specification): when a point lies on a mirror line, its image is unchanged; when it does not, reflect by keeping perpendicular distance to the mirror line. This supports accurate reflection descriptions in coordinate geometry.