Transformations (Mappings) of Figures

A transformation of the plane is a rule that assigns to each point of the plane exactly one image point. Transformations are the fundamental tool for comparing figures and defining the key geometric notions of congruence and similarity.

Core Definitions

📐Definition — Transformation (Mapping) of the Plane

A transformation (or mapping) of the plane is a rule $f$ that assigns to every point $A$ of the plane a unique point $A' = f(A)$ , called the image of $A$ . The original point $A$ is called the preimage of $A'$ .

📐Definition — Bijection

A transformation $f$ is called a bijection (one-to-one and onto) if every point of the plane is the image of exactly one preimage. Equivalently:

Injective: $f(A) = f(B)$ implies $A = B$ (distinct points have distinct images).
Surjective: every point $P$ of the plane has some preimage, i.e.\ there exists $A$ with $f(A) = P$ .

All geometric transformations studied in school geometry (translations, reflections, rotations, homotheties) are bijections.

Isometries (Motions)

📐Definition — Isometry (Motion)

A bijection $f$ of the plane is called an isometry (or motion) if it preserves the distance between every pair of points:

$\boxed{|A'B'| = |AB| \quad \text{for all points } A, B \text{ of the plane}}$

where $A' = f(A)$ and $B' = f(B)$ .

Because an isometry preserves all distances, it automatically preserves:

Angles — the angle between two lines equals the angle between their images.
Shape and size — the image of any figure is congruent to the original.
Collinearity — three collinear points map to three collinear points.
Betweenness — if $B$ lies between $A$ and $C$ , then $B'$ lies between $A'$ and $C'$ .

The four main types of isometries are: parallel translation, axial symmetry (reflection), central symmetry, and rotation.

Direct and Opposite Isometries

📐Definition — Direct and Opposite Isometries

An isometry is called direct (or orientation-preserving) if it maps every figure to a congruent figure with the same orientation — i.e.\ a figure traversed counterclockwise maps to a figure also traversed counterclockwise.

An isometry is called opposite (or orientation-reversing) if it reverses orientation — a figure traversed counterclockwise maps to one traversed clockwise.

Isometry	Orientation
Parallel translation	Direct
Rotation	Direct
Central symmetry	Direct
Axial symmetry (reflection)	Opposite

A practical test: if the image of a figure is a mirror image (like a left glove becoming a right glove), the transformation is opposite; otherwise it is direct.

Worked Examples

✎Example — Example 1 — Identifying the Transformation

Triangles $ABC$ and $A'B'C'$ have vertices $A(1,0)$ , $B(4,0)$ , $C(4,3)$ and $A'(-1,0)$ , $B'(-4,0)$ , $C'(-4,-3)$ . Identify which transformation maps $\triangle ABC$ to $\triangle A'B'C'$ and determine whether it is direct or opposite.

Solution. Compare corresponding coordinates:

$A(1,0) \to A'(-1,0), \quad B(4,0) \to B'(-4,0), \quad C(4,3) \to C'(-4,-3).$

Each image point is obtained by negating both coordinates: $(x,y) \to (-x,-y)$ . This is the central symmetry with center at the origin.

Check distances: $|AB| = 3$ , $|A'B'| = 3$ ✓; $|BC| = 3$ , $|B'C'| = 3$ ✓; $|CA| = \sqrt{9+9} = 3\sqrt{2}$ , $|C'A'| = 3\sqrt{2}$ ✓. So $f$ is an isometry.

Orientation: the vertices $A, B, C$ go counterclockwise; $A', B', C'$ also go counterclockwise (check: the signed area of $A'B'C'$ has the same sign as $A'B'C'$ traversed in the listed order). Central symmetry is a direct isometry. $\blacktriangleleft$

✎Example — Example 2 — Verifying an Isometry

Is the mapping $f(x, y) = (x + 2,\; |y|)$ an isometry of the plane?

Solution. Take $A = (0, 1)$ and $B = (0, -1)$ . Then $A' = (2, 1)$ and $B' = (2, 1)$ , so $A' = B'$ .

But $A \neq B$ , which contradicts injectivity. Therefore $f$ is not a bijection and hence not an isometry. (Alternatively: $|AB| = 2$ but $|A'B'| = 0 \neq 2$ .) $\blacktriangleleft$

Exercises

✏Exercise

Points $P(2, 5)$ and $Q(-3, 1)$ are given. A transformation maps every point $(x, y)$ to $(y, x)$ (reflection across the line $y = x$ ). Find the images $P'$ and $Q'$ and verify that $|P'Q'| = |PQ|$ .

✏Exercise

Determine whether each of the following mappings is an isometry. Justify your answer.

(a) $f(x, y) = (x + 3,\; y - 1)$

(b) $g(x, y) = (2x,\; 2y)$