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The vertices of ABCD are A(2, 8), B(2, 4), C(8, 4), and D(8, 8). Draw ABCD. Find the image A1B1C1D1 of ABCD under a reflection in the y-axis. Find the image A2B2C2D2 of A1B1C1D1 under a reflection in

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2 parts
b

A1(−2,8),B1(−2,4),C1(−8,4),D1(−8,8)A_1(-2, 8), B_1(-2, 4), C_1(-8, 4), D_1(-8, 8)

A2(−2,−6),B2(−2,−2),C2(−8,−2),D2(−8,−6)A_2(-2, -6), B_2(-2, -2), C_2(-8, -2), D_2(-8, -6)

d

A3(−2,6),B3(−2,2),C3(−8,2),D3(−8,6)A_3(-2, 6), B_3(-2, 2), C_3(-8, 2), D_3(-8, 6)

Step-by-step solution

12 steps

Here's the solution for question 6.

Part a

The vertices of ABCDABCD are given as A(2,8)A(2, 8), B(2,4)B(2, 4), C(8,4)C(8, 4), and D(8,8)D(8, 8).

To draw ABCDABCD, plot these points on a graph sheet with perpendicular axes OxOx and OyOy, using a scale of 2 cm2 \text{ cm} to 2 units2 \text{ units} (which means 1 cm1 \text{ cm} represents 1 unit1 \text{ unit}) for intervals −10≤x≤10-10 \le x \le 10 and −10≤y≤10-10 \le y \le 10. Then connect the points AA to BB, BB to CC, CC to DD, and DD to AA.

Part b

A1(−2,8),B1(−2,4),C1(−8,4),D1(−8,8)A_1(-2, 8), B_1(-2, 4), C_1(-8, 4), D_1(-8, 8)

To find the image A1B1C1D1A_1B_1C_1D_1 of ABCDABCD under a reflection in the y-axis, we use the transformation rule (x,y)→(−x,y)(x, y) \to (-x, y).

  1. Apply the rule to point $A(2, 8)$.

    A1=(−2,8)A_1 = (-2, 8)

  2. Apply the rule to point $B(2, 4)$.

    B1=(−2,4)B_1 = (-2, 4)

  3. Apply the rule to point $C(8, 4)$.

    C1=(−8,4)C_1 = (-8, 4)

  4. Apply the rule to point $D(8, 8)$.

    D1=(−8,8)D_1 = (-8, 8)

    The vertices of A1B1C1D1A_1B_1C_1D_1 are A1(−2,8),B1(−2,4),C1(−8,4),D1(−8,8)\boxed{A_1(-2, 8), B_1(-2, 4), C_1(-8, 4), D_1(-8, 8)}.

    c) To find the image A2B2C2D2A_2B_2C_2D_2 of A1B1C1D1A_1B_1C_1D_1 under a reflection in the line y=1y=1, we use the transformation rule (x,y)→(x,2k−y)(x, y) \to (x, 2k - y), where k=1k=1. So the rule is (x,y)→(x,2(1)−y)=(x,2−y)(x, y) \to (x, 2(1) - y) = (x, 2 - y).

  5. Apply the rule to point A1(−2,8)A_1(-2, 8).

    A2=(−2,2−8)=(−2,−6)A_2 = (-2, 2 - 8) = (-2, -6)

  6. Apply the rule to point B1(−2,4)B_1(-2, 4).

    B2=(−2,2−4)=(−2,−2)B_2 = (-2, 2 - 4) = (-2, -2)

  7. Apply the rule to point C1(−8,4)C_1(-8, 4).

    C2=(−8,2−4)=(−8,−2)C_2 = (-8, 2 - 4) = (-8, -2)

  8. Apply the rule to point D1(−8,8)D_1(-8, 8).

    D2=(−8,2−8)=(−8,−6)D_2 = (-8, 2 - 8) = (-8, -6)

    The vertices of A2B2C2D2A_2B_2C_2D_2 are A2(−2,−6),B2(−2,−2),C2(−8,−2),D2(−8,−6)\boxed{A_2(-2, -6), B_2(-2, -2), C_2(-8, -2), D_2(-8, -6)}.

Part d

A3(−2,6),B3(−2,2),C3(−8,2),D3(−8,6)A_3(-2, 6), B_3(-2, 2), C_3(-8, 2), D_3(-8, 6)

To find the image A3B3C3D3A_3B_3C_3D_3 of A2B2C2D2A_2B_2C_2D_2 under a reflection in the x-axis, we use the transformation rule (x,y)→(x,−y)(x, y) \to (x, -y).

  1. Apply the rule to point A2(−2,−6)A_2(-2, -6).

    A3=(−2,−(−6))=(−2,6)A_3 = (-2, -(-6)) = (-2, 6)

  2. Apply the rule to point B2(−2,−2)B_2(-2, -2).

    B3=(−2,−(−2))=(−2,2)B_3 = (-2, -(-2)) = (-2, 2)

  3. Apply the rule to point C2(−8,−2)C_2(-8, -2).

    C3=(−8,−(−2))=(−8,2)C_3 = (-8, -(-2)) = (-8, 2)

  4. Apply the rule to point D2(−8,−6)D_2(-8, -6).

    D3=(−8,−(−6))=(−8,6)D_3 = (-8, -(-6)) = (-8, 6)

    The vertices of A3B3C3D3A_3B_3C_3D_3 are A3(−2,6),B3(−2,2),C3(−8,2),D3(−8,6)\boxed{A_3(-2, 6), B_3(-2, 2), C_3(-8, 2), D_3(-8, 6)}.

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