1.8 Congruence by Rigid Motions — Transformations

Standards: G-CO.B.6

A fabrication shop cuts two metal plates from the same digital outline. One plate is moved, turned, and flipped on the inspection table so its edges can be compared with the other plate.

The inspector also studies the plate by itself. Some turns or flips place the outline exactly onto its starting outline, while other movements do not.

What movement evidence could show that the two plates have the same size and shape, and what movements could reveal symmetry in one plate?

Given two figures or a transformation sequence, you will identify a rigid-motion mapping or a nonrigid step and justify the congruence conclusion.

Definitions and rules

Rigid motions preserve congruence

If a sequence made only of translations, rotations, and reflections carries one figure onto another, then the figures are congruent.

A dilation establishes congruence only when its scale factor is 1.

Why it has to be this: Each rigid motion preserves every distance and angle measure, so a composition of rigid motions preserves them as well.

Properties preserved by common transformations.
Transformation	Distances	Angles	Orientation	Congruence
Translation	Preserved	Preserved	Preserved	Preserved
Rotation	Preserved	Preserved	Preserved	Preserved
Reflection	Preserved	Preserved	Reversed	Preserved
Dilation with k = 1	Preserved	Preserved	Preserved	Preserved
Dilation with k > 0 and k ≠ 1	Scaled	Preserved	Preserved	Not preserved

Worked examples

Example 1 — quadrilaterals matched by a reflection and translation

Quadrilateral ABCD has vertices A(−4, 1), B(−1, 1), C(0, 3), and D(−3, 4). Quadrilateral A′B′C′D′ has vertices A′(2, −2), B′(−1, −2), C′(−2, 0), and D′(1, 1). Decide whether the figures are congruent and identify a mapping sequence.

Suggested work

Givens

A(−4, 1) B(−1, 1) C(0, 3) D(−3, 4) A′(2, −2) B′(−1, −2) C′(−2, 0) D′(1, 1)

Tools

Reflect over the y-axis: (x, y) → (−x, y) Then apply T⟨−2, −3⟩

Primitives

Use the same two transformations on every vertex.

Computation

A(−4, 1) → (4, 1) → (2, −2) = A′ B(−1, 1) → (1, 1) → (−1, −2) = B′ C(0, 3) → (0, 3) → (−2, 0) = C′ D(−3, 4) → (3, 4) → (1, 1) = D′ Both transformations are rigid motions, so the quadrilaterals are congruent.

Notice: The same mapping sequence must work for every corresponding vertex.

Why coordinate agreement is enough

The sequence maps every named vertex to its corresponding image, and rigid motions preserve each segment between those vertices.

The transformed boundary therefore occupies the target boundary with every length and angle preserved.

Example 2 — a rotation followed by a size change

Triangle PQR with vertices P(1, 1), Q(3, 1), and R(1, 2) is rotated 90° counterclockwise, then dilated about the origin by k = 2. Is the final image congruent to the preimage?

Suggested work

Givens

P(1, 1) Q(3, 1) R(1, 2) R90, then D(2)

Tools

R90(x, y) = (−y, x) D(2)(x, y) = (2x, 2y)

Primitives

Original PQ = 2 Track PQ through both transformations.

Computation

P(1, 1) → P′(−1, 1) → P′′(−2, 2) Q(3, 1) → Q′(−1, 3) → Q′′(−2, 6) R(1, 2) → R′(−2, 1) → R′′(−4, 2) P′′Q′′ = 4, while PQ = 2. The dilation changes length, so the final image is not congruent to the preimage.

Notice: One nonrigid transformation prevents the full composition from certifying congruence.

Example 3 — a triangle moved by one translation

Triangle UVW has vertices U(−2, 0), V(1, 0), and W(0, 3). Apply T⟨−5, 2⟩ and decide whether the image is congruent to the preimage.

Suggested work

Givens

U(−2, 0) V(1, 0) W(0, 3) T⟨−5, 2⟩

Tools

T⟨a, b⟩(x, y) = (x + a, y + b)

Primitives

a = −5 b = 2

Computation

U(−2, 0) → U′(−7, 2) V(1, 0) → V′(−4, 2) W(0, 3) → W′(−5, 5) A translation is a rigid motion, so triangle UVW is congruent to triangle U′V′W′.

Common traps

Orientation is not congruence

A reflection reverses clockwise and counterclockwise order, but it preserves every length and angle. Reversed orientation does not make the reflected image noncongruent.

Practice

A figure is translated and then reflected. Does this sequence preserve congruence? Name the evidence used.
Answer — tap to show
Yes. Both transformations are rigid motions, so their composition preserves every distance and angle measure.
Suggested work — tap to show
Givens
translation, then reflection
Tools
translation = rigid motion reflection = rigid motion
Primitives
Check each transformation in the sequence.
Computation
rigid motion → rigid motion → congruence preserved
A figure is dilated by k = 3/2 and then rotated. Can this sequence establish that the final figure is congruent to the original?
Answer — tap to show
No. The dilation scales nonzero lengths by 3/2, and the rotation preserves the changed lengths rather than restoring the original lengths.
Suggested work — tap to show
Givens
D(3/2), then a rotation
Tools
A rotation is rigid. A dilation with k ≠ 1 is not rigid.
Primitives
Track one nonzero side length L.
Computation
L → (3/2)L → (3/2)L The final length differs from L.
A figure is reflected over the x-axis. Are the image and preimage congruent?
Answer — tap to show
Yes. A reflection is a rigid motion.
A polygon is dilated by k = 1/2. Can the dilation alone carry the polygon onto a congruent image when the polygon has a nonzero side length?
Answer — tap to show
No. Every nonzero side length is multiplied by 1/2, so the image has a different size.
Suggested work — tap to show
Givens
D(1/2) side length L > 0
Tools
image length = kL
Primitives
k = 1/2
Computation
L → (1/2)L Since L > 0, (1/2)L ≠ L.
Triangle ABC has vertices A(−3, 2), B(−1, 4), and C(2, 1). Triangle A′B′C′ has vertices A′(−1, −1), B′(1, −3), and C′(4, 0). Find a rigid-motion sequence that maps the first triangle to the second and state the congruence conclusion.
Answer — tap to show
Reflect over the x-axis, then translate by T⟨2, 1⟩. Therefore triangle ABC is congruent to triangle A′B′C′.
Suggested work — tap to show
Givens
A(−3, 2) B(−1, 4) C(2, 1) A′(−1, −1) B′(1, −3) C′(4, 0)
Tools
Reflect over x-axis: (x, y) → (x, −y) T⟨2, 1⟩
Primitives
Apply both transformations to every vertex.
Computation
A(−3, 2) → (−3, −2) → (−1, −1) = A′ B(−1, 4) → (−1, −4) → (1, −3) = B′ C(2, 1) → (2, −1) → (4, 0) = C′ Reflection and translation are rigid motions, so congruence is preserved.