Transformations — Solutions

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1. Translations

   1. A(1,2) + (3,4): A′(4, 6)
   2. B(5,−1) + (−2,3): B′(3, 2)
   3. C(−3,4) + (5,−2): C′(2, 2)
   4. D(0,−6) + (−4,6): D′(−4, 0)
   5. E(7,7) + (−7,−7): E′(0, 0)
   6. F(−2,−5) + (6,8): F′(4, 3)
   7. G(4,0) + (0,−3): G′(4, −3)
   8. H(−1,3) + (−3,−5): H′(−4, −2)

2. Reflections

   1. A(3,5) in x-axis: A′(3, −5)
   2. B(−2,4) in y-axis: B′(2, 4)
   3. C(6,−3) in x-axis: C′(6, 3)
   4. D(−4,−1) in y-axis: D′(4, −1)
   5. E(2,2) in y=x: E′(2, 2) — invariant point (lies on the line)
   6. F(5,−3) in y=x: F′(−3, 5)
   7. G(4,1) in x=2: G′(0, 1) — 4 is 2 units right of x=2, so image is 2 units left: x=0
   8. H(1,5) in y=3: H′(1, 1) — 5 is 2 above y=3, so image is 2 below: y=1

3. Rotations about the origin

   1. A(3,1) 90° anticlockwise: A′(−1, 3)
   2. B(4,−2) 90° clockwise: B′(−2, −4)
   3. C(−1,5) 180°: C′(1, −5)
   4. D(2,3) 90° clockwise: D′(3, −2)
   5. E(−3,−4) 90° anticlockwise: E′(4, −3)
   6. F(0,6) 180°: F′(0, −6)
   7. G(5,0) 90° anticlockwise: G′(0, 5)
   8. H(−2,4) 90° clockwise: H′(4, 2)

4. Describe the transformation

   1. (1,1)→(4,3): Translation by vector (3, 2) — 3 right and 2 up
   2. (1,2)→(1,−2): Reflection in the x-axis
   3. (2,1)→(−2,1): Reflection in the y-axis
   4. (0,2)→(2,0): Rotation of 90° clockwise about the origin
   5. (1,0)→(0,−1): Rotation of 90° clockwise about the origin

5. Composition of transformations

   1. P(3,2) translate then reflect x-axis: After translation: (5, −1). After reflection in x-axis: P′′(5, 1)
   2. Q(−1,4) reflect y-axis then translate: After reflection: (1, 4). After translation: Q′′(4, 5)
   3. Triangle translate then rotate 90° CW: After translation by (−4,2): A′(−3,2), B′(−1,2), C′(−1,4). After 90° CW rotation (x,y)→(y,−x): A′′(2,3), B′′(2,1), C′′(4,1)
   4. R(2,5) rotate 180° then reflect x-axis: After 180°: (−2,−5). After reflect in x-axis: (−2,5). Equivalent to reflection in the y-axis.

6. Problem solving

   1. Tiling floor:
      1. Translate by vector (2, 0) — 2 right
      2. Translate by vector (0, 2) — 2 up
      3. Row 3, column 2: start at (0,0), add (2,0) once for col 2 and (0,2) twice for row 3. Vertices: (2,4), (4,4), (4,6), (2,6)
   2. Logo reflections:
      1. Reflect in y-axis: A′(−1,1), B′(−4,1), C′(−4,5)
      2. Reflect in x=5: distance from x=5: A is 4 right (x=1), so image at x=9; B at x=4 is 1 right, image at x=6; C at x=4 image at x=6. A′(9,1), B′(6,1), C′(6,5)
   3. Mirror line: After reflection and a 6-unit right translation, the shape returns to its original position. The translation moves it back 6 units. The mirror line is x = 3 (midway, so reflection moves it 6 units left, then translation moves it 6 right).
   4. Congruent images: All three transformations are isometries — they preserve distances and angles. Side lengths and angle sizes are unchanged, so the image is always congruent to the pre-image.

7. Triangle T — multiple transformations

   1. Rotate 90° anticlockwise (x,y)→(−y,x): A(2,1)→A′(−1,2); B(5,1)→B′(−1,5); C(5,4)→C′(−4,5). T′ has vertices (−1,2), (−1,5), (−4,5).
   2. Reflect in y=x (swap x and y): A(2,1)→A′′(1,2); B(5,1)→B′′(1,5); C(5,4)→C′′(4,5). T′′ has vertices (1,2), (1,5), (4,5).
   3. Translate by (−3,2) then reflect in y-axis: After translation: A→(−1,3), B→(2,3), C→(2,6). After reflect in y-axis (x→−x): A′′′(1,3), B′′′(−2,3), C′′′(−2,6).
   4. Which transformation gives all vertices in Q2 (x<0, y>0): Part (a): T′ has vertices (−1,2), (−1,5), (−4,5) — all x negative, all y positive. This is the 90° anticlockwise rotation.

8. Equivalent single transformations

   1. Reflect P(3,5) in x-axis twice: First reflection: (3,−5). Second reflection: (3,5) — back to original. Equivalent to the identity transformation (no change).
   2. Rotate Q(4,2) 90° ACW twice (= 180°): After first 90° ACW: (−2,4). After second: (−4,−2). Equivalent to a single 180° rotation about the origin.
   3. Translate R(1,3) by (4,−2) then (−4,2): Net vector: (4−4, −2+2) = (0,0). Equivalent to the identity — the point returns to (1,3). No movement.
   4. Reflect S(2,−1) in x-axis then y-axis: After x-axis reflection: (2,1). After y-axis reflection: (−2,1). Equivalent to a 180° rotation about the origin.

9. Invariant points and symmetry

   1. Invariant points under reflection in x-axis: Any point (x, 0) — all points on the x-axis. When y = 0, the reflection maps (x, 0) to (x, 0).
   2. Invariant under rotation about origin: Only the origin (0, 0) — the centre of rotation is always invariant.
   3. Reflect P(a,b) in y=x, maps to itself when: Reflection in y=x maps (a,b) to (b,a). For this to equal (a,b), we need b=a. So P maps to itself when a = b (i.e., when P lies on the line y=x).
   4. Rotations mapping square onto itself: Rotations about centre (2,2): 90° clockwise, 90° anticlockwise, 180°, and 360° (identity). Four rotations in total map the square onto itself.

10. Design and reasoning

    1. Combined translation vector: Add the vectors: (3+(−5), 4+1) = (−2, 5). Single translation by vector (−2, 5).
    2. Quadrilateral A, B, B′, A′: After reflecting in y-axis: A′(−1,0), B′(−3,0). The four points are A(1,0), B(3,0), B′(−3,0), A′(−1,0). These are collinear (all on the x-axis, y=0), so they form a degenerate case — actually a line segment, since C(2,3) is the apex. The full shape A, B, C, B′, A′ forms a pentagon. Just A, B, B′, A′ forms a trapezium (one pair of parallel sides on the x-axis with A′A parallel to B′B).
    3. Tile translation vectors: (i) To the right: vector (3,0) — the width of the tile. (ii) Immediately above: vector (1,2) — from base to top-left corner. (iii) 2 right and 1 up: vector (3×2+1×1, 0×2+2×1) = (7,2).
    4. Two reflections in parallel lines = translation: Example: reflect P(1,3) in x=0 (y-axis) to get P′(−1,3), then reflect in x=3 to get P′′(7,3). The result is a translation of 6 units right — twice the distance between the parallel lines (0 and 3, separation = 3, translation = 2×3 = 6 in the direction from the first line to the second).