Transformations and Vectors

Transformations

Reflection

The shape is flipped over a mirror line (like the $x$-axis, the $y$-axis or any line like $x = 2$ or $y = -1$). The perpendicular distance of a point from the line of reflection remains the same.

As demonstrated in the question above, you have to make sure to keep the perpendicular distance of the new shape from the line of reflection the same as the original shape.

To make things easier, memorise the table below that shows how the coordinates change when you reflect a shape.

Rotation

This is like turning a shape around a point, called the centre of rotation, by a certain angle and direction (clockwise or anticlockwise).

These rules turn the shape about the origin. Memorise the table.

In this question, $(0, 1)$ is the centre of rotation. Each corner stays the same distance from $(0, 1)$ and turns a quarter turn clockwise around it. You can check a corner with the rule: move the centre to the origin (subtract $(0, 1)$), apply $(x, y) \to (y, -x)$, then move back (add $(0, 1)$). For the corner $(-3, 5)$:

$(-3, 5) \to (-3, 4) \to (4, 3) \to (4, 4)$

Translation

A translation is written like this: $\begin{pmatrix} a \\ b \end{pmatrix}$.

• $a$ $=$ how many steps right ($+$) or left ($-$).
• $b$ $=$ how many steps up ($+$) or down ($-$).

Every corner moves 4 right and 1 down.

Enlargement

This changes the size of the shape, but not the shape itself. It can grow or shrink based on the scale factor $k$.

It should also be noted that the shape changes from the perspective of a centre of enlargement. The table above shows the change in coordinates if the centre of enlargement is the origin $(0, 0)$.

Draw a line from the origin through each corner, and go twice as far along it. The corners $(1, 3)$, $(1, 1)$ and $(2, 1)$ become $(2, 6)$, $(2, 2)$ and $(4, 2)$.

A negative scale factor puts the image on the opposite side of the centre, and $\frac{1}{2}$ makes it half the size. Take each corner's position relative to $(-1, 2)$, multiply by $-\frac{1}{2}$, and add $(-1, 2)$ back. For the corner $(1, 5)$:

$(1, 5) - (-1, 2) = (2, 3) \quad\to\quad -\tfrac{1}{2}(2, 3) = (-1, -1.5) \quad\to\quad (-2, 0.5)$

The other corners $(1, 1)$ and $(3, 1)$ become $(-2, 2.5)$ and $(-3, 2.5)$.

Vectors

We have already learned how vectors are used to perform translations on shapes and points. Now, let's learn a little more about them.

Addition and subtraction

You can add and subtract vectors by adding or subtracting the top numbers and the bottom numbers:

$\begin{pmatrix} 7 \\ 2 \end{pmatrix} + \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \end{pmatrix} \qquad \begin{pmatrix} 7 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$

Multiplication with a scalar

You can also multiply an integer (a scalar) with a vector. Multiply both numbers:

$-3\begin{pmatrix} 5 \\ -3 \end{pmatrix} = \begin{pmatrix} -15 \\ 9 \end{pmatrix}$

Magnitude of a vector

Magnitude means "how long" the vector is: its length or size. The formula for finding the magnitude of a vector is:

$|\mathbf{v}| = \sqrt{3^2 + 4^2} = 5$

Vector geometry

Vectors can be represented using arrows. For example, the vector $\overrightarrow{OA} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ (from the origin to the point $A$) can be represented on the graph as:

If two vectors are scalar multiples of each other, they are parallel. For example, if $\overrightarrow{OA} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix} 1 \\ 1.5 \end{pmatrix}$, then they are parallel (because $\overrightarrow{OA} = 2\overrightarrow{BC}$).

Three points $A$, $B$ and $C$ are collinear (fall on the same line) if the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel. Because the two vectors share the point $A$, being parallel is enough to put all three points on one straight line.

Here $\overrightarrow{AB} = \mathbf{a}$ and $\overrightarrow{AC} = 2\mathbf{a}$. They point in the same direction, so $A$, $B$ and $C$ are collinear.

Some questions may ask you to calculate vectors based on ratios.

We can deduce from the diagram that $\overrightarrow{OC} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}$. The ratio says that $\overrightarrow{OB}$ is one part for every 3 parts of $\overrightarrow{OC}$, so we divide $\overrightarrow{OC}$ by 3 to get $\overrightarrow{OB}$:

$\overrightarrow{OB} = \tfrac{1}{3}\overrightarrow{OC} = \tfrac{1}{3}\begin{pmatrix} 6 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$