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

Circle the best answer.
Kira claims that reflecting across y = x and then turning 90° counterclockwise about the origin gives the same result in either order. Test it on D(3, 1). Which verdict is right?
- Right: both orders send D to (−3, 1)
- Wrong: a reflection then a turn is not a rigid motion
- Wrong: D lands on (−3, 1) one way and (3, −1) the other
- Right: both moves keep lengths, so the two results must match
Triangle A(1, 1), B(4, 1), C(1, 3) must land on A′(−1, −1), B′(−4, −1), C′(−1, −3), 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 A′B′C′ are drawn with their corners labeled. Coordinate grid, all four quadrants, each square one unit, origin in the middle. Triangles ABC and A′B′C′ are drawn with their corners labeled. x y origin each square one unit A B C A' B' C' - Reflect across the y-axis, then across y = x
- Reflect across y = x, then across y = −x
- A half-turn about the origin
- Reflect across the x-axis, then across the y-axis
Noah maps quadrilateral ABCD onto PQRS by a dilation with factor 2 and then a rotation, and says the shapes are congruent. Which verdict is right?
Quadrilateral ABCD and a larger quadrilateral PQRS are sketched; the dilation and the rotation are written beside them. Quadrilateral ABCD and a larger quadrilateral PQRS are sketched; the dilation and the rotation are written beside them. A B C D dilation rotation P Q R S - Shown: the shapes have the same angles, so they are congruent
- Not shown: a dilation with factor 2 doubles every length
- Shown: the rotation is rigid, so the two shapes are congruent
- Not shown: the rotation changes the angles of ABCD
Tala reflects a figure across the x-axis and then across y = x. The lines meet at the origin at 45°. Which single motion does the same?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. The x-axis and the line y = x are dashed. Coordinate grid, all four quadrants, each square one unit, origin in the middle. The x-axis and the line y = x are dashed. −10 −8 −6 −4 −2 2 4 6 8 10 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10 x-axis y = x - A turn of 45° counterclockwise about the origin
- A reflection across the y-axis
- A turn of 90° counterclockwise about the origin
- A turn of 90° clockwise about the origin
Ari wants two reflections in lines through the origin that together turn a figure 120° counterclockwise. Which pair works?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. Lines through the origin are named by their angle from the positive x-axis. Coordinate grid, all four quadrants, each square one unit, origin in the middle. Lines through the origin are named by their angle from the positive x-axis. −10 −8 −6 −4 −2 2 4 6 8 10 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10 Lines through the origin are named by their angle from the positive x-axis. - The x-axis, then the line at 60°
- The x-axis, then the y-axis
- The line at 60°, then the x-axis
- The x-axis, then the line at 120°
A reflection maps P(1, 5) to P′(5, 1). Which line is the mirror line, and why does Bo know?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. P(1, 5) and P′(5, 1) are marked; no line is drawn. Coordinate grid, all four quadrants, each square one unit, origin in the middle. P(1, 5) and P′(5, 1) are marked; no line is drawn. −10 −8 −6 −4 −2 2 4 6 8 10 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10 P P' - y = −x + 6: the line through P and P′
- y = x: through the midpoint (3, 3), at right angles to PP′
- y = −x: it is at right angles to PP′
- x = 3: the vertical line through the midpoint (3, 3) of PP′
A rigid motion maps A(0, 0) to A′(2, 1) and B(3, 4) to B′. Which point does Esme find could be B′?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. A(0, 0), B(3, 4) and A′(2, 1) are marked. Coordinate grid, all four quadrants, each square one unit, origin in the middle. A(0, 0), B(3, 4) and A′(2, 1) are marked. −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 0 A B A' - (4, 5)
- (7, 2)
- (6, 4)
- (5, 4)
Lines ℓ and m are perpendicular and meet at P. Jun turns both 30° about a point O, not P. What must be true of the images ℓ′ and m′?
Sketch: lines ℓ ⊥ m meeting at P, and a separate point O; the turn is 30° counterclockwise. Sketch: lines ℓ ⊥ m meeting at P, and a separate point O; the turn is 30° counterclockwise. l m 30° counterclockwise P O - They meet at 30° plus 90°, 120°
- They are perpendicular and meet at P′, the image of P
- They are perpendicular just when O is P
- They are parallel to ℓ and m and still meet at the point P
AB = DE, AC = DF and ∠A = ∠D. Kira slides A onto D, then turns until ray AB lies on ray DE. Why does C then land on F?
Sketch: triangles ABC and DEF with AB = DE, AC = DF and ∠A = ∠D marked. Sketch: triangles ABC and DEF with AB = DE, AC = DF and ∠A = ∠D marked. A B C D E F - C lands on F once BC = EF is checked as well, as in SSS
- C lands on F because the two triangles have equal areas
- C lands on F, or on F's mirror image in DE; a reflection in DE then fixes it
- C must land on F, since rigid motions keep lengths and angles
Luz slides parallelogram ABCD 5 units right and then reflects it across the x-axis. Which statement about the image A′B′C′D′ is true?
Coordinate grid, all four quadrants, each square one unit, origin in the middle. Parallelogram ABCD is drawn above the x-axis with its corners labeled. Coordinate grid, all four quadrants, each square one unit, origin in the middle. Parallelogram ABCD is drawn above the x-axis with its corners labeled. A B C D 1 unit x y - Opposite sides stay parallel and equal, but the corners run the other way
- Its corners run in the same order as ABCD, as a slide keeps order
- Its sides are each 5 units longer, stretched by the slide
- Its opposite sides may no longer be parallel after the reflection