Transformations and Similarity in Geometry

In coordinate geometry, we distinguish between the original shape, called the object, and the transformed shape, known as the image. When certain points do not change their position following a transformation, they are referred to as invariant points. This page details the four primary 2-dimensional transformations and how they affect the properties of shapes.

Rotation

A 2-dimensional rotation turns an object about a fixed point in the plane. To define a rotation completely, three pieces of information are required:

1. The centre of rotation: the fixed point about which the shape turns.
2. The angle of rotation: the degree of turning between the object and image.
3. The direction of rotation: by convention, anticlockwise is the positive direction.

Under rotation, the object and image are exactly the same size and shape, meaning they are congruent. The only invariant point is the centre of rotation, unless the rotation is $0^{\circ}$ or a multiple of $360^{\circ}$, in which case every point on the plane is invariant.

Reflection

A 2-dimensional reflection involves flipping an object across a fixed line, known as the reflection line or mirror line. The object and image remain congruent, but because the shape must be 'turned over' to match the image, they are said to have opposite congruence. The set of invariant points for a reflection is the mirror line itself: every point on the line remains exactly where it is.

Translation and Translation Vectors

A translation moves an object a specified distance in a specified direction without any rotation or reflection. This movement is described by the number of units the object shifts parallel to the $x$ and $y$ axes. We express this using a 2-dimensional vector form:

$\begin{pmatrix} a \\ b \end{pmatrix}$

In this vector, $a$ represents the horizontal displacement (the $x$-direction). A positive $a$ indicates a move to the right, while a negative $a$ indicates a move to the left. The value $b$ represents the vertical displacement (the $y$-direction). A positive $b$ indicates a move upwards, and a negative $b$ indicates a move downwards.

Translations preserve both the size and orientation of the shape, so the object and image are congruent. There are no invariant points in a translation, unless the vector is the zero vector $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$, in which case all points are invariant.

Enlargement

An enlargement transforms an object based on a centre of enlargement and a scale factor $n$. The image is $n$ times the size of the object and is located $n$ times as far from the centre of enlargement as the object. Properties of the scale factor include:

1. Negative scale factors: These indicate that the object and image are on opposite sides of the centre of enlargement.
2. Fractional scale factors: If the scale factor is between $-1$ and $1$, the image is smaller than the object and located closer to the centre.
3. Congruence vs Similarity: Enlargement preserves shape but changes size. Therefore, the object and image are similar rather than congruent. They are only congruent if the scale factor is $1$ or $-1$.

The centre of enlargement is the only invariant point, unless the scale factor is $1$, which leaves all points invariant.

Combinations of Transformations

Multiple transformations can be applied in sequence. It is vital to perform them in the specific order given, as changing the order often leads to a different final image. Any combination consisting only of rotations, reflections, and translations will always result in an image congruent to the original object.

Worked Example: Rotation

A square has vertices $A(1, 0)$, $B(3, 0)$, $C(3, 2)$, and $D(1, 2)$. It is rotated $90^{\circ}$ anticlockwise about the centre $(5, 5)$. What are the image coordinates $A'$, $B'$, $C'$, and $D'$?

Method 1: Counting Squares Always draw a diagram. You can determine the position of $A'$ by looking at the displacement from the centre $(5, 5)$. Since $A$ is $5$ squares down and $4$ squares left from the centre, a $90^{\circ}$ anticlockwise rotation means $A'$ will be $4$ squares down and $5$ squares right from the centre. This places $A'$ at $(10, 1)$. Repeating this logic for the other vertices gives $B'(10, 3)$, $C'(8, 3)$, and $D'(8, 1)$.

Method 2: Using the Centre of the Shape For symmetrical shapes like squares or circles, you can rotate the centre of the shape itself and then reconstruct the shape around that new point.

Worked Example: Reflection

A trapezium with vertices $A(0, 0)$, $B(3, 0)$, $C(2, 2)$, and $D(0, 2)$ is reflected in the line $x + y = 5$. To find the images $A'$, $B'$, $C'$, and $D'$, draw the mirror line and the object. For each vertex, draw a line at right angles to the mirror line and extend it the same distance on the other side. This results in $A'(5, 5)$, $B'(5, 2)$, $C'(3, 3)$, and $D'(3, 5)$.

Worked Example: Translation

A trapezium with vertices $A(1, 6)$, $B(4, 6)$, $C(3, 8)$, and $D(1, 8)$ is moved $6$ squares right and $4$ squares down. We can express this as the translation vector $\begin{pmatrix} 6 \\ -4 \end{pmatrix}$. Applying this shift to each vertex gives $A'(7, 2)$, $B'(10, 2)$, $C'(9, 4)$, and $D'(7, 4)$.

Worked Example: Enlargement

Consider a triangle with vertices $A(2, 2)$, $B(4, 2)$, and $C(2, 6)$ and centre of enlargement $P(0, 4)$.

1. Scale factor 2: Join $P$ to each vertex and double the distance. For $A$, the vector $PA = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$. The image $A'$ is found via $OA' = OP + 2PA = \begin{pmatrix} 0 \\ 4 \end{pmatrix} + 2\begin{pmatrix} 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}$, so $A'$ is $(4, 0)$.
2. Scale factor $\frac{1}{2}$: The distance from $P$ is halved, keeping the image on the same side of $P$.
3. Scale factor -2: The distance is doubled, but the image is projected through $P$ to the opposite side. This results in an inverted triangle.

Combined Transformations and Invariance

Example 1: Trapezium $T$ (vertices as in the translation example) is reflected in $y = 4$ to form $P$, then $P$ is reflected in $x = 5$ to form $Q$. By inspection, the single transformation mapping $T$ to $Q$ is a rotation of $180^{\circ}$ about the centre $(5, 4)$. The point $(5, 4)$ is the only invariant point.

Example 2: Trapezium $T$ with vertices $A(2, 7)$, $B(5, 7)$, $C(4, 9)$, and $D(2, 9)$ is translated by $\begin{pmatrix} 5 \\ -1 \end{pmatrix}$ to form $P$, and $P$ is then rotated $90^{\circ}$ clockwise about $(4, 6)$ to form $Q$. To find the single transformation from $T$ to $Q$, we observe it is a $90^{\circ}$ clockwise rotation. The centre of this rotation is found by constructing the perpendicular bisectors of the lines connecting $A$ to $A'$ and $D$ to $D'$. The intersection is $(1, 4)$, which is the invariant point.