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What single transformation maps 0394A 2B 2C 2 onto 0394A 3B 3C 3?

Mathematics

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Step-by-step solution

12 steps

a) I cannot draw the graph for you, but you should draw two perpendicular axes, OxOx and OyOy, on graph paper. The xx-axis should range from −10-10 to 10, and the yy-axis from −12-12 to 12. Use a scale where 2 cm2 \text{ cm} represents 2 units2 \text{ units} on both axes.

Part b(i)

The vertices of triangle ABCABC are given as

I cannot draw the triangles, but I will provide the coordinates for each vertex.

$A(4, 4)$
$B(6, 8)$
$C(2, 8)$

Part b(ii)· 3 steps

To find the image △A1B1C1\triangle A_1B_1C_1 of △ABC\triangle ABC under a reflection in the line x=−2x = -2, we use the transformation rule (x,y)→(2k−x,y)(x, y) \to (2k - x, y), where k=−2k = -2.

So, (x,y)→(−4−x,y)(x, y) \to (-4 - x, y).

  1. Apply the reflection rule to point AA.

    A(4,4)→A1(−4−4,4)=A1(−8,4)A(4, 4) \to A_1(-4 - 4, 4) = A_1(-8, 4)

  2. Apply the reflection rule to point BB.

    B(6,8)→B1(−4−6,8)=B1(−10,8)B(6, 8) \to B_1(-4 - 6, 8) = B_1(-10, 8)

  3. Apply the reflection rule to point CC.

    C(2,8)→C1(−4−2,8)=C1(−6,8)C(2, 8) \to C_1(-4 - 2, 8) = C_1(-6, 8)

    The vertices of △A1B1C1\triangle A_1B_1C_1 are A1(−8,4),B1(−10,8),C1(−6,8)\boxed{A_1(-8, 4), B_1(-10, 8), C_1(-6, 8)}.

Part b(iii)· 3 steps

To find the image △A2B2C2\triangle A_2B_2C_2 of △ABC\triangle ABC under an enlargement with scale factor −1-1 from the origin, we use the transformation rule (x,y)→(kx,ky)(x, y) \to (kx, ky), where k=−1k = -1.

So, (x,y)→(−x,−y)(x, y) \to (-x, -y).

  1. Apply the enlargement rule to point AA.

    A(4,4)→A2(−4,−4)A(4, 4) \to A_2(-4, -4)

  2. Apply the enlargement rule to point BB.

    B(6,8)→B2(−6,−8)B(6, 8) \to B_2(-6, -8)

  3. Apply the enlargement rule to point CC.

    C(2,8)→C2(−2,−8)C(2, 8) \to C_2(-2, -8)

    The vertices of △A2B2C2\triangle A_2B_2C_2 are A2(−4,−4),B2(−6,−8),C2(−2,−8)\boxed{A_2(-4, -4), B_2(-6, -8), C_2(-2, -8)}.

Part b(iv)· 3 steps

To find the image △A3B3C3\triangle A_3B_3C_3 of △ABC\triangle ABC under a reflection in the line y=0y = 0 (the xx-axis), we use the transformation rule (x,y)→(x,2k−y)(x, y) \to (x, 2k - y), where k=0k = 0.

So, (x,y)→(x,−y)(x, y) \to (x, -y).

  1. Apply the reflection rule to point AA.

    A(4,4)→A3(4,−4)A(4, 4) \to A_3(4, -4)

  2. Apply the reflection rule to point BB.

    B(6,8)→B3(6,−8)B(6, 8) \to B_3(6, -8)

  3. Apply the reflection rule to point CC.

    C(2,8)→C3(2,−8)C(2, 8) \to C_3(2, -8)

    The vertices of △A3B3C3\triangle A_3B_3C_3 are A3(4,−4),B3(6,−8),C3(2,−8)\boxed{A_3(4, -4), B_3(6, -8), C_3(2, -8)}.

Part c· 3 steps

To find the single transformation that maps △A2B2C2\triangle A_2B_2C_2 onto △A3B3C3\triangle A_3B_3C_3, we compare their coordinates

A2(−4,−4)A_2(-4, -4) maps to A3(4,−4)A_3(4, -4)
B2(−6,−8)B_2(-6, -8) maps to B3(6,−8)B_3(6, -8)
C2(−2,−8)C_2(-2, -8) maps to C3(2,−8)C_3(2, -8)

  1. Observe the change in coordinates from A2A_2 to A3A_3.

    The yy-coordinate remains the same (e.g., −4→−4-4 \to -4), while the xx-coordinate changes sign (e.g., −4→4-4 \to 4).

  2. Confirm this pattern for B2→B3B_2 \to B_3 and C2→C3C_2 \to C_3.

    For BB: (−6,−8)→(6,−8)(-6, -8) \to (6, -8). x→−xx \to -x, y→yy \to y.
    For CC: (−2,−8)→(2,−8)(-2, -8) \to (2, -8). x→−xx \to -x, y→yy \to y.

  3. Identify the transformation.

    The transformation rule is (x,y)→(−x,y)(x, y) \to (-x, y). This is a reflection in the yy-axis (the line x=0x=0).
    The single transformation is a reflection in the y-axis\boxed{\text{reflection in the } y\text{-axis}}.

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