Geometry Rigid Transformations Worksheet Answer Key

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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers

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Looking for a rigid transformations worksheet answer key? Then this answer key gives every step of the Geometry worksheet on Translations, Reflections and Rotations Explained. Each solution applies a coordinate rule, shows the image points and checks a length or a picture when helpful. Compare the order in which motions are applied, because the final image can depend on it.

Standards: G-CO.A.2, G-CO.A.3, G-CO.A.4, G-CO.A.5 · Level: Grade 10 Geometry

Problem 1 Translate a point

Subtract 5 from the x-coordinate and add 4 to the y-coordinate: (3 − 5, −2 + 4) = (−2, 2).

P′(−2, 2)

Problem 2 Reflect over an axis

Over the x-axis the sign of y changes: (−4, −6). Over the y-axis the sign of x changes: (4, 6).

(a) (−4, −6), (b) (4, 6)

Problem 3 Half-turn about the origin

A 180° rotation changes both signs: (−7, 3).

R′(−7, 3)

Problem 4 Quarter-turn rotation

Use (x, y) → (−y, x): (−(−1), 4) = (1, 4).

S′(1, 4)

Problem 5 Reflect over y = x

Reflecting over y = x swaps the coordinates: (5, −2).

T′(5, −2)

Problem 6 Describe a translation

The vector goes from A to A′: ⟨2 − (−3), −2 − 1⟩ = ⟨5, −3⟩. Apply it to the other vertices: B′ = (0 + 5, 1 − 3) = (5, −2) and C′ = (−2 + 5, 4 − 3) = (3, 1).

Triangle and its translated image.

Vector ⟨5, −3⟩; B′(5, −2); C′(3, 1)

Problem 7 Rigid or not rigid

The segment goes to (4, −1)–(6, −1), still length 2: rigid (a translation).
The segment goes to (0, 0)–(4, 0), length 4: not rigid (a dilation).
The segment is unchanged in length, and the rule is a reflection: rigid.
The segment goes to (0, 0)–(6, 0), length 6: not rigid (a horizontal stretch).

(a) rigid, (b) not rigid, (c) rigid, (d) not rigid

Problem 8 Reflect a quadrilateral

Negate each x-coordinate: W′(−2, 1), X′(−6, 2), Y′(−7, 5), Z′(−3, 4).

Preimage and image are mirror images across the y-axis.

W′(−2, 1), X′(−6, 2), Y′(−7, 5), Z′(−3, 4)

Problem 9 Clockwise rotation

A 90° clockwise rotation is a 270° counterclockwise rotation: (x, y) → (y, −x). D′(1, −2), E′(1, −6), F′(4, −6).

Rotated image in the fourth quadrant.

D′(1, −2), E′(1, −6), F′(4, −6)

Problem 10 Reflect over y = −x

For P: (−(−5), −3) = (5, −3). For Q: (−(−6), −(−2)) = (6, 2).

P′(5, −3), Q′(6, 2)

Problem 11 Symmetries of a rectangle

Reflections: over the x-axis and over the y-axis, so two lines of symmetry. The diagonals are not lines of symmetry because the sides are unequal.

Rotations: only 180° about the origin, since a 90° turn would swap the 6-unit and 4-unit sides.

A square has four lines of symmetry (two axes, two diagonals) and rotations of 90°, 180° and 270°.

Rectangle: 2 lines, 180° rotation; square: 4 lines, rotations of 90°, 180°, 270°

Problem 12 Regular polygon symmetry

A regular n-gon has n lines of symmetry and a smallest rotation of 360° ÷ n.

Polygon	Lines of symmetry	Smallest rotation
Pentagon	5	360 ÷ 5 = 72°
Octagon	8	360 ÷ 8 = 45°
12-gon	12	360 ÷ 12 = 30°

Pentagon 5 lines, 72°; octagon 8 lines, 45°; 12-gon 12 lines, 30°

Problem 13 Order of two motions

Translate first: (1, −5); then reflect over the x-axis: (1, 5).

Reflect first: (−7, −4); then translate: (1, −13).

The two final points differ, so the order matters.

(1, 5) versus (1, −13)

Problem 14 Find the line of reflection

The line of reflection is the perpendicular bisector of AA′. The points differ only in y, so the line is horizontal through the midpoint: y = (5 + (−1)) ÷ 2 = 2.

Reflecting over y = 2 keeps x and sends y to 2(2) − y. For B: 4 − (−3) = 7, so B′ = (6, 7).

Line y = 2; B′(6, 7)

Problem 15 Robot docking route

After the translation: (−3 + 7, 2 − 4) = (4, −2). Reflecting over the x-axis gives (4, 2).

The first move has length √(7² + 4²) = √65 ≈ 8.1 feet.

Dock at (4, 2); first move ≈ 8.1 ft

Problem 16 Rotation about another center

Move the center to the origin by subtracting (2, 1): P becomes (3, 2).

90° counterclockwise: (3, 2) → (−2, 3). Add (2, 1) back: (0, 4).

180°: (3, 2) → (−3, −2). Add (2, 1) back: (−1, −1).

Check: CP = √(9 + 4) = √13, and the distance from C to (0, 4) is √(4 + 9) = √13.

(0, 4) and (−1, −1)

Problem 17 Reflection then translation

Reflection over the y-axis: A′(−1, 1), B′(−4, 1), C′(−4, 3).

Translation by ⟨0, −5⟩: A″(−1, −4), B″(−4, −4), C″(−4, −2).

Check: AB = 4 − 1 = 3 and A″B″ = |−4 − (−1)| = 3.

Final image after both motions.

A″(−1, −4), B″(−4, −4), C″(−4, −2); both lengths are 3

Problem 18 Identify the rotation

Test the rule (x, y) → (−y, x): P(1, 2) → (−2, 1) ✓, Q(4, 2) → (−2, 4) ✓, R(4, 4) → (−4, 4) ✓.

The horizontal side PQ of length 3 became the vertical side P′Q′ of length 3, which fits a quarter turn.

A 90° counterclockwise rotation about the origin: (x, y) → (−y, x)

Problem 19 Mosaic with four-fold symmetry

Apply (x, y) → (−y, x) repeatedly: (3, 1) → (−1, 3) → (−3, −1) → (1, −3), and one more turn returns to (3, 1).

The reflection over y = x swaps coordinates, so it maps (3, 1) to (1, 3).

(3, 1), (−1, 3), (−3, −1), (1, −3); the line y = x

Problem 20 Two reflections make a translation

Reflection over x = 1 sends (x, y) to (2 − x, y). Over x = 4 it sends (x, y) to (8 − x, y).

For P: (0, 2) → (2, 2) → (6, 2). The motion is the translation ⟨6, 0⟩, which is twice the distance 3 between the lines.

General point: (x, y) → (2 − x, y) → (8 − (2 − x), y) = (x + 6, y).

Final point (6, 2); the translation ⟨6, 0⟩

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