Rigid Motions and Transformations: practice solutions, Grade 10 – download the PDF

1 Translating points ★★★

The rule is \((x, y)\to(x - 4,\; y + 5)\).

1. \(P′ = (3 - 4,\; -2 + 5) = (-1, 3)\).
2. \(Q′ = (0 - 4,\; 6 + 5) = (-4, 11)\).

2 Three reflections ★★★

1. \((x, y)\to(x, -y)\), so \(A′(4, -7)\).
2. \((x, y)\to(-x, y)\), so \(B′(3, 2)\).
3. \((x, y)\to(y, x)\), so \(C′(5, 5)\). The point does not move because it lies on the line \(y = x\), and points on the line of reflection are fixed.

3 Turning about the origin ★★★

1. \((x, y)\to(-y, x)\): \(A′(-1, 3)\).
2. \((x, y)\to(-x, -y)\): \(B′(2, -4)\).
3. \((x, y)\to(y, -x)\): \(C′(-2, -5)\).

4 True or false? ★★★

1. False. A translation preserves distances, so the size never changes.
2. True. If the vertices are read counterclockwise on the preimage, they are read clockwise on the image.
3. True. A full turn brings every point back to its starting position.
4. False. A dilation with scale factor 2 doubles every length, so distances are not preserved.

5 Counting symmetries ★★★

1. 2 lines; \(180^\circ\).
2. 3 lines; \(360^\circ \div 3 = 120^\circ\).
3. 6 lines; \(360^\circ \div 6 = 60^\circ\).
4. 1 line (through the midpoints of the parallel sides); none.
5. 0 lines; \(180^\circ\).

6 Finding the translation ★★★

1. \(\langle -1 - 2,\; 9 - 5\rangle = \langle -3, 4\rangle\).
2. \(B′ = (0 - 3,\; 0 + 4) = (-3, 4)\).
3. Undo the move: \((7 + 3,\; -2 - 4) = (10, -6)\).

7 Lengths do not change ★★★

\(PQ = \sqrt{(5 - 1)^2 + (5 - 2)^2} = \sqrt{16 + 9} = 5\).

\(P′ = (4, -2)\) and \(Q′ = (8, 1)\), so \(P′Q′ = \sqrt{4^2 + 3^2} = 5\).

The two lengths are equal, as they must be for a rigid motion.

8 A vertical mirror line ★★★

A point at horizontal distance \(d\) from the line goes to the other side at the same distance, so \(x\) becomes \(2\cdot 3 - x = 6 - x\).

\((7, 2)\to(-1, 2)\); \((1, -4)\to(5, -4)\); \((3, 6)\to(3, 6)\) (on the line).

Rule: \((x, y)\to(6 - x,\; y)\).

9 A horizontal mirror line ★★★

The x-coordinate stays the same. The y-coordinate is reflected about \(-1\): its new value is \(2(-1) - y = -2 - y\).

\((2, 3)\to(2, -5)\); \((-4, -1)\to(-4, -1)\), fixed because it is on the line; \((0, -6)\to(0, 4)\).

10 Reflecting a triangle in y = x ★★★

Swap the coordinates: \(A′(4, 1)\), \(B′(0, 3)\), \(C′(2, 5)\).

Area of \(ABC\) by the shoelace formula: \(\frac12\left|1(0 - 2) + 3(2 - 4) + 5(4 - 0)\right| = \frac12|-2 - 6 + 20| = 6\).

Area of \(A′B′C′\): \(\frac12\left|4(3 - 5) + 0(5 - 1) + 2(1 - 3)\right| = \frac12|-12| = 6\).

The areas are equal, which is expected for a rigid motion.

11 Order matters ★★★

1. \(T\): \((6, 2)\). Then \(M\): \((-6, 2)\).
2. \(M\): \((-4, -1)\). Then \(T\): \((-2, 2)\).

The results \((-6, 2)\) and \((-2, 2)\) are different, so the order of the moves matters.

12 A clockwise quarter turn ★★★

A clockwise quarter turn is \((x, y)\to(y, -x)\).

\(D′(0, -2)\), \(E′(1, -4)\), \(F′(3, -3)\).

The same image comes from a rotation of \(270^\circ\) counterclockwise.

13 Drone on a grid map ★★★

1. x: \(2 + 4 - 6 + 1 = 1\). y: \(3 + 1 + 2 - 7 = -1\). The drone finishes at \((1, -1)\).
2. Add the vectors: \(\langle 4 - 6 + 1,\; 1 + 2 - 7\rangle = \langle -1, -4\rangle\). Check: \((2 - 1,\; 3 - 4) = (1, -1)\).

It finishes 1 meter left and 4 meters down from where it started.

14 Are these triangles congruent? ★★★

1. \(AB = 6\), \(AC = 2\), \(BC = \sqrt{36 + 4} = \sqrt{40}\). \(DE = 6\), \(DF = 2\), \(EF = \sqrt{4 + 36} = \sqrt{40}\). The three pairs of sides are congruent (SSS), so the triangles are congruent.
2. Translate by \(\langle 1, 1\rangle\): \(A\to(1, 1) = D\), \(B\to(7, 1)\), \(C\to(1, 3)\). Then rotate \(90^\circ\) counterclockwise about \(D(1, 1)\): \(B\) is 6 units right of \(D\), so it goes 6 units above \(D\), at \((1, 7) = E\). \(C\) is 2 units above \(D\), so it goes 2 units left of \(D\), at \((-1, 1) = F\).

15 Two parallel mirrors ★★★

1. Reflection in \(x = 1\): \(x\to 2 - x\), so \(P\to(4, 3)\). Reflection in \(x = 5\): \(x\to 10 - x\), so \((4, 3)\to(6, 3)\).
2. In general \(x\to 10 - (2 - x) = x + 8\) and \(y\) is unchanged: the translation by \(\langle 8, 0\rangle\). The lines are 4 units apart and \(2\times 4 = 8\). Test: \((0, -2)\to(2, -2)\to(8, -2)\), which is \((0 + 8, -2)\).

16 Two mirrors that meet ★★★

1. \((x, y)\to(x, -y)\to(-x, -y)\): a rotation of \(180^\circ\) about the origin. The axes meet at \(90^\circ\) and \(2\times 90^\circ = 180^\circ\).
2. \((x, y)\to(y, x)\to(-y, x)\): this is the \(90^\circ\) counterclockwise rotation about the origin. The lines meet at \(45^\circ\) and \(2\times 45^\circ = 90^\circ\). Test: \((3, 4)\to(4, 3)\to(-4, 3)\), and the rule gives \((-4, 3)\).

17 Finding the mirror line ★★★

1. The line is the perpendicular bisector of \(\overline{AA′}\). Midpoint: \(\left(\frac{1 + 5}{2}, \frac{6 + 2}{2}\right) = (3, 4)\). Slope of \(AA′\): \(\frac{2 - 6}{5 - 1} = -1\), so the mirror line has slope \(1\). Through \((3, 4)\): \(y - 4 = x - 3\), that is \(y = x + 1\).
2. \(B(0, 1)\): \(0 + 1 = 1\), so \(B\) is on the line and is not moved.

18 Glide reflection and orientation ★★★

1. Reflection: \((-4, -2)\), \((-1, -2)\), \((-3, -5)\). Translation: \(R″(2, -1)\), \(S″(5, -1)\), \(T″(3, -4)\).
2. Before: \(-4(2 - 5) - 1(5 - 2) - 3(2 - 2) = 12 - 3 + 0 = 9\). After: \(2(-1 + 4) + 5(-4 + 1) + 3(-1 + 1) = 6 - 15 + 0 = -9\). The absolute value is the same (area \(4.5\)), but the sign changes: the orientation has been reversed by the reflection.

19 Rotating about another center ★★★

Subtract \(C\), rotate about the origin, add \(C\).

1. \(P - C = (3, 3)\to(-3, 3)\to(-1, 4)\).
2. \(Q - C = (0, -4)\to(0, 4)\to(2, 5)\).
3. \(R - C = (4, 0)\to(0, -4)\to(2, -3)\).

20 Why reflections keep distances ★★★

\(P′(-a, b)\) and \(Q′(-c, d)\). Then \(P′Q′^2 = (-c + a)^2 + (d - b)^2 = (a - c)^2 + (b - d)^2 = PQ^2\), so \(P′Q′ = PQ\).

Check: \(PQ = \sqrt{3^2 + 4^2} = 5\). \(P′(-2, 3)\), \(Q′(1, 7)\): \(P′Q′ = \sqrt{3^2 + 4^2} = 5\).

21 Symmetry detective ★★★

1. 8 lines; \(360^\circ \div 8 = 45^\circ\).
2. \(360^\circ \div 5 = 72^\circ\), so the angles are \(72^\circ, 144^\circ, 216^\circ, 288^\circ, 360^\circ\).
3. \(n = 360 \div 24 = 15\): a regular 15-gon (15 sides).

22 The Ferris wheel ★★★

1. \(360^\circ \div 12 = 30^\circ\).
2. \(210 \div 30 = 7\) positions, so car 5 reaches position \(5 + 7 = 12\).
3. \(90 \div 30 = 3\) positions: \(10 + 3 = 13\), which wraps around to position 1.