Transformations & Vectors

Transformations: the big four

A transformation moves or changes a shape on a coordinate grid. At Higher tier you must describe, perform, and combine four transformations. The original shape is the object; the result is the image. We usually label image points with a dash, so $A$ maps to $A'$.

Key terms Congruent — same shape and size. Translation, reflection and rotation all produce congruent images.

Similar — same shape, different size. Enlargement produces a similar image (unless the scale factor is $\pm 1$).

Invariant point — a point that does not move under the transformation.

A golden rule for the exam: one transformation, one description. If a question gives one mark for the type and the rest for the details, naming two transformations (e.g. "reflection and rotation") usually scores zero.

Translation

A translation slides every point the same distance in the same direction, with no turning or resizing. It is described by a column vector:

$$\begin{pmatrix} x \\ y \end{pmatrix}$$

where $x$ is the movement right (or left if negative) and $y$ is the movement up (or down if negative).

To translate a shape, add the vector to every vertex. For example, translating $A(2,1)$ by $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ gives $A'(5,-1)$.

Exam tip To find the translation vector between two shapes, pick one vertex on the object and the matching vertex on the image, then do image − object. Read "across then up" to avoid sign slips.

Reflection

A reflection flips the shape over a mirror line. Every image point is the same perpendicular distance from the line as its object point. To describe a reflection fully you must give the equation of the mirror line.

Common mirror lines:

Rotation

A rotation turns the shape about a fixed centre of rotation. To describe it fully you need three things:

1. The angle (usually $90°$, $180°$ or $270°$).
2. The direction (clockwise or anticlockwise) — though for $180°$ direction does not matter.
3. The centre (a coordinate).

Exam tip Use tracing paper if the exam allows it: trace the object, press a pencil tip on the centre, and turn. To find a centre, draw the perpendicular bisectors of two lines joining object–image point pairs; they cross at the centre.

A quick check for $90°$ about the origin: anticlockwise sends $(x,y)\to(-y,x)$; clockwise sends $(x,y)\to(y,-x)$.

Enlargement

An enlargement changes size by a scale factor $k$ from a centre of enlargement. Lengths multiply by $k$; area multiplies by $k^2$.

$k > 1$: image is larger.
$0 < k < 1$ (fractional): image is smaller.
$k < 0$ (negative): image is on the opposite side of the centre and turned upside down.

To enlarge, draw a ray from the centre through each vertex. Multiply the distance (in grid steps) from the centre to each vertex by $k$. For negative $k$, measure the same multiple but in the opposite direction through the centre.

Worked example Enlarge triangle $P(1,1)$, $Q(1,2)$, $R(3,1)$ by scale factor $-2$, centre $(0,0)$.

Combining transformations

A combined transformation applies one transformation, then another to the result. The exam often asks you to describe the single transformation equivalent to two others.

Useful facts:

Two reflections in parallel lines give a translation.
Two reflections in intersecting lines give a rotation about the intersection point.
A $180°$ rotation about the origin is the same as an enlargement by scale factor $-1$.

Watch out Order matters. "Reflect in $y=x$ then translate" usually gives a different image from "translate then reflect". Carry out the first transformation fully, plot the image, then apply the second to that image.

Vectors: column form

A vector has both magnitude (size) and direction. We write it as a column vector:

$$\overrightarrow{AB} = \begin{pmatrix} x \\ y \end{pmatrix}$$

To find a vector between points, do end − start. The vector from $A(1,2)$ to $B(4,7)$ is $\begin{pmatrix} 4-1 \\ 7-2 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$.

Add or subtract vectors component by component:

$$\begin{pmatrix} 3 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}$$

A scalar multiple stretches a vector: $3\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}$. The result is parallel to the original.

The magnitude (length) uses Pythagoras:

$$\left| \begin{pmatrix} x \\ y \end{pmatrix} \right| = \sqrt{x^2 + y^2}$$

So $\left|\begin{pmatrix} 3 \\ 4 \end{pmatrix}\right| = \sqrt{9+16} = 5$.

Key terms Parallel vectors — one is a scalar multiple of the other: $\mathbf{a}$ is parallel to $\mathbf{b}$ if $\mathbf{a} = k\mathbf{b}$.

Collinear points — three or more points on the same straight line. Show two vectors along the line are parallel and share a common point.

Vector geometry proofs

These questions use point notation like $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. Build every path by walking along known vectors:

$$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$$

Worked example $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $AB$. Show $\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$.

To prove collinear, show that two vectors are scalar multiples and share a point. If $\overrightarrow{XY} = 2\mathbf{m}$ and $\overrightarrow{XZ} = 3\mathbf{m}$, then $\overrightarrow{XY}$ and $\overrightarrow{XZ}$ are parallel and both start at $X$, so $X$, $Y$, $Z$ are collinear.

Exam tip Always finish a proof with a written sentence: "the vectors are parallel and share the common point $X$, therefore the points are collinear." The final mark is for the conclusion, not just the algebra.