Reflect over any line and rotate about a point, with coordinate rules
Grade 10 Math Worksheet - Challenge Me

Circle the best answer.
A rotation maps A(3, 1) to A′(1, 3) and B(4, 2) to B′(0, 4). Where is the center, and why does Milo know?
- (2, 2): the midpoint of AA′
- (2, 3): the midpoint of BB′
- (1, 1): on the perpendicular bisectors of AA′ and BB′
- (0, 0): rotations on this grid turn about the origin
Noah reflects a figure across y = x and then across x + y = 2. Which single motion does the same?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. The lines y = x and x + y = 2 are dashed; they meet at (1, 1). Coordinate grid, all four quadrants, each square one unit, origin in the middle. The lines y = x and x + y = 2 are dashed; they meet at (1, 1). −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 0 y = x x + y = 2 - A slide along the line y = x
- A half-turn about (1, 1)
- A turn of 90° about (1, 1)
- A half-turn about the origin
Pia reflects a figure across y = x and then across y = x + 4. These lines are parallel. Which single slide does the same?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. The parallel lines y = x and y = x + 4 are dashed. Coordinate grid, all four quadrants, each square one unit, origin in the middle. The parallel lines y = x and y = x + 4 are dashed. −6 −4 −2 2 4 6 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10 y = x y = x + 4 - (−2, 2): the gap between the lines
- (4, −4): twice the gap, but back toward the first line
- (0, 8): twice the 4-unit vertical gap
- (−4, 4): twice the 2√2 gap, across the lines
A reflection maps P(−1, 4) to P′(5, 2). Which line does Tala find is the mirror line?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. P(−1, 4) and P′(5, 2) are marked; no line is drawn. Coordinate grid, all four quadrants, each square one unit, origin in the middle. P(−1, 4) and P′(5, 2) are marked; no line is drawn. −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 −5 −4 −3 −2 −1 1 2 3 4 5 6 0 P P' - y = 3x − 3
- x = 2, through the midpoint
- y = 3x, through the origin
- y = −x/3 + 11/3, through P and P′
A rotation about C maps A to A′. Zee says C must lie on the perpendicular bisector of AA′. Which reason is right?
Sketch: points A and A′, segment AA′, and a center C off the segment. Sketch: points A and A′, segment AA′, and a center C off the segment. A A' C - AA′ passes through C, which is fixed by the turn
- CA = CA′, since a turn keeps distance to its center
- The rotation turns AA′ by 90° about its own midpoint
- C is the midpoint of AA′, so it splits AA′ in half
A half-turn maps A(−2, 1) to C(6, 3) and B(1, −1) to D(3, 5). Where is its center, and what does Ari conclude about ABCD?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. A, B, C and D are marked and joined in order. Coordinate grid, all four quadrants, each square one unit, origin in the middle. A, B, C and D are marked and joined in order. A B C D - (2, 2); ABCD is a rectangle
- (−0.5, 0), the midpoint of AB; ABCD is a parallelogram
- (0, 0), the origin; ABCD is a parallelogram
- (2, 2), the midpoint of both AC and BD; a parallelogram
A rotation maps A(0, 4) to A′(4, 0). Which point does Bo show could NOT be its center?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. A(0, 4) and A′(4, 0) are marked. Coordinate grid, all four quadrants, each square one unit, origin in the middle. A(0, 4) and A′(4, 0) are marked. −5 −4 −3 −2 −1 1 2 3 4 5 6 7 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 0 A A' - (2, 2)
- (5, 5)
- (3, 1)
- (0, 0)
Two lines of symmetry of a regular pentagon meet at its center at 36°. Jun reflects across one and then the other. What single motion results?
Sketch: a regular pentagon with two of its lines of symmetry drawn, 36° apart at the center. Sketch: a regular pentagon with two of its lines of symmetry drawn, 36° apart at the center. 36° - A reflection across a third line of symmetry of the pentagon
- A turn of 72°, which maps the pentagon onto itself
- A turn of 36° about the center
- A turn of 144° about the center
“each square one unit” Triangle A(0, 0), B(4, 0), C(0, 3) must land on P(2, 1), Q(2, 5), R(−1, 1), corner to corner. Which sequence does Luz find fails?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. Triangles ABC and PQR are drawn with their corners labeled. Coordinate grid, all four quadrants, each square one unit, origin in the middle. Triangles ABC and PQR are drawn with their corners labeled. x y 0 A B C P Q R - Turn 90° counterclockwise about (0.5, 1.5)
- Turn 90° counterclockwise about P(2, 1)
- Turn 90° counterclockwise about the origin, then slide 2 right and 1 up
- Slide 2 right and 1 up, then turn 90° counterclockwise about P
Pia reflects across the x-axis and then turns 90° counterclockwise about the origin, and claims the result is one reflection across y = x. Which verdict is right?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. Point (3, 1) is marked to test the claim. Coordinate grid, all four quadrants, each square one unit, origin in the middle. Point (3, 1) is marked to test the claim. −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 0 - Wrong: a reflection and then a turn make a turn
- Right: (x, y) → (x, −y) → (y, x)
- Wrong: it is a 90° turn about the origin
- Wrong: it is a reflection across y = −x