Transformations in Geometry

A geometric transformation assigns an image point to each original point. Applying one rule to every point of a figure produces its image. The central questions are what moves, what stays fixed and which properties are preserved.

The original figure is often called the preimage. A prime names the corresponding image: $A$ maps to $A’$ , and segment $AB$ maps to $A’B’$ . Correspondence lets you compare the same feature before and after the transformation rather than two unrelated parts of a drawing.

On this page

Four familiar transformations

A translation moves every point by the same vector. A rotation turns every point through the same directed angle about one centre. A reflection places each point the same perpendicular distance on the other side of a mirror line. A dilation scales every displacement from one centre by a common non-zero factor.

The rule must specify enough information to determine the image. “Move right” lacks a distance; “rotate $90^\circ$ ” lacks a centre and direction; “enlarge by $2$ ” lacks a centre. A fully stated rule can be applied without guessing from the picture.

Changing the order changes the endpoint

T moves two units right; R turns 90° counterclockwise about the origin. R(T(P)) and T(R(P)) differ.

Diagram explanation and exact data

The solid route goes (1,0) → (3,0) → (0,3). The dashed route goes (1,0) → (0,1) → (2,1). The arcs indicate the actual rotation paths.

Changing the order changes the endpoint: exact reference data
Order	Intermediate point	Endpoint
T then R	$(3,0)$	$(0,3)$
R then T	$(0,1)$	$(2,1)$

Rigid motions preserve size

Translations, rotations and reflections are isometries, or rigid motions. For any two original points $P,Q$ , their images satisfy $P’Q’=PQ$ . This distance preservation also preserves angles, perimeter, area, straightness and parallelism.

A composition of rigid motions is again rigid: preserving a distance at every stage preserves it overall. This gives a transformation definition of congruence, as developed in AMSI’s Congruence module.

Reflection reverses clockwise versus counterclockwise vertex order. Translation and rotation preserve that order. Reversed orientation does not mean a figure has changed size or lost congruence.

Dilation preserves shape, not usually size

For a dilation of signed factor $k\ne0$ , corresponding distances satisfy $P’Q’=|k|PQ$ . Angles are preserved, while area is multiplied by $k^2$ . If $|k|\ne1$ , the image is similar but not congruent to the original.

A negative $k$ puts an image on the opposite ray through the centre. In a plane this is a positive dilation followed by a $180^\circ$ rotation, not a reflection in a line. A factor of zero collapses every point to the centre, so it is excluded from the invertible dilations considered here.

Dilation preserves shape, not usually size: reference table
Transformation	Data required	Length factor	Typical fixed points
Translation	A vector	$1$	None for a non-zero vector
Rotation	Centre and directed angle	$1$	The centre for a non-full-turn rotation
Reflection	A mirror line	$1$	Every point on the line
Dilation	Centre and non-zero factor	$\|k\|$	The centre when $k\ne1$

The identity transformation leaves every point fixed. It appears as translation by the zero vector, rotation by a full turn or dilation by $1$ .

Working with a complete coordinate rule

Translate a triangle with $A(1,1)$ , $B(4,1)$ and $C(2,3)$ by vector $(3,-2)$ . Add the same displacement to each point:

A’=(4,-1),\qquad B’=(7,-1),\qquad C’=(5,1).

The original base vector is $B-A=(3,0)$ and the image base vector is $B’-A’=(3,0)$ . Its length remains $3$ . The same cancellation happens for any pair of points, so this is not merely an isolated numerical coincidence.

Compare a dilation of the original triangle by factor $2$ about the origin. The vertices become $(2,2)$ , $(8,2)$ and $(4,6)$ . Its base length is $6$ , twice the original. The original area is $3$ square units; the image area is $12$ , four times as large.

Order can change the result

Let $T$ translate right by $2$ , and let $R$ rotate $90^\circ$ counterclockwise about the origin. Starting from $P(1,0)$ :

R(T(P))=R(3,0)=(0,3).

In the other order,

T(R(P))=T(0,1)=(2,1).

The two results differ. When transformations are written as a composition $R\circ T$ , the rightmost rule, $T$ , acts first. A verbal instruction “translate, then rotate” follows the same order.

Two translations commute because vector addition commutes. Rotations about the same centre also combine by adding angles. Those special cases do not make all transformations commutative.

Recovering a transformation from evidence

To identify a translation, subtract each original point from its image and check that the same vector appears. To identify a rotation, look for one common centre and one common directed angle, with centre-to-point distances unchanged. For reflection, the mirror line bisects every non-trivial point-image segment at right angles.

Equal corresponding lengths establish congruence of appropriate figures but may not uniquely identify the transformation used. A symmetric figure can match its image in several ways. Naming corresponding labelled points removes some of that ambiguity.

A square distorted to a non-square rectangle is not a dilation. It may preserve parallel sides, but the unequal horizontal and vertical scale factors do not preserve all angles or shape.

Practice questions

Translate $(-2,4)$ by $(5,-3)$ .
Is a dilation of factor $3$ an isometry? What happens to an area of $7$ square units?
Apply “right $2$ , then $90^\circ$ counterclockwise about the origin” to $(1,0)$ .

Check your answers

Add coordinatewise: $(-2+5,4-3)=(3,1)$ .
No. Lengths triple, so distances are not preserved. The area becomes $3^2(7)=63$ square units.
The intermediate point is $(3,0)$ and the final point is $(0,3)$ . Reversing the order would give $(2,1)$ .

Sources and further reading

AMSI, Congruence: translations, rotations and reflections as congruence transformations.
AMSI, Scale Drawings and Similarity: uniform scale and similar figures.