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Difficulty: Very hardTransformations in the Coordinate Plane

In the standard (x,y)(x,y) coordinate plane, triangle PQRPQR has vertices P(2,3)P(2, 3), Q(6,3)Q(6, 3), and R(2,6)R(2, 6). The triangle undergoes a sequence of three transformations:

1. A dilation centered at the point (4,2)(4, 2) with a scale factor of −2-2.
2. A reflection across the line y=−xy = -x.
3. A rotation of 90∘90^\circ counterclockwise about the origin.

What are the coordinates of the image of vertex RR after this sequence of transformations?

  1. A
    (6,−8)(6, -8)
  2. B
    (−4,12)(-4, 12)
  3. (8,6)(8, 6)Answer
  4. D
    (−8,−6)(-8, -6)
  5. E
    (0,−10)(0, -10)

Answer

The coordinate pair (8,6)(8, 6)
Applying the dilation formula centered at (4,2)(4, 2) with scale factor −2-2 to R(2,6)R(2, 6) gives the point R′(8,−6)R'(8, -6). Reflecting this point across the line y=−xy = -x negates and swaps the coordinates, yielding R′′(6,−8)R''(6, -8). Finally, a 90∘90^\circ counterclockwise rotation about the origin swaps the coordinates and negates the new xx-coordinate, producing the final coordinates (8,6)(8, 6).

Step-by-Step Solution

1
Perform the dilation of vertex R(2,6)R(2, 6) centered at C(4,2)C(4, 2) with a scale factor of k=−2k = -2.
The intermediate image is R′(8,−6)R'(8, -6).
For a dilation centered at (xc,yc)(x_c, y_c) with scale factor kk, the coordinates of the image are (xc+k(x−xc),yc+k(y−yc))(x_c + k(x - x_c), y_c + k(y - y_c)). Substituting R(2,6)R(2, 6), C(4,2)C(4, 2), and k=−2k = -2 gives: x′=4−2(2−4)=8x' = 4 - 2(2 - 4) = 8 and y′=2−2(6−2)=−6y' = 2 - 2(6 - 2) = -6.
2
Reflect the point R′(8,−6)R'(8, -6) across the line y=−xy = -x.
The intermediate image is R′′(6,−8)R''(6, -8).
A reflection across the line y=−xy = -x maps any point (x,y)(x, y) to (−y,−x)(-y, -x). Applying this rule to R′(8,−6)R'(8, -6) yields R′′(−(−6),−8)=(6,−8)R''(-(-6), -8) = (6, -8).
3
Rotate the point R′′(6,−8)R''(6, -8) by 90∘90^\circ counterclockwise about the origin.
The final image is R′′′(8,6)R'''(8, 6).
A rotation of 90∘90^\circ counterclockwise about the origin maps any point (x,y)(x, y) to (−y,x)(-y, x). Applying this rule to R′′(6,−8)R''(6, -8) yields R′′′(−(−8),6)=(8,6)R'''(-(-8), 6) = (8, 6).

Key Concept

Composite transformations in the coordinate plane combining dilation from a non-origin center, reflection across diagonal lines, and rotation about the origin.

Alternative Method

Instead of applying the transformations step-by-step to the point, we can track the transformations vectorially. For the dilation, the vector CR⃗=R−C=(−2,4)\vec{CR} = R - C = (-2, 4) is scaled by −2-2 to get −2CR⃗=(4,−8)-2\vec{CR} = (4, -8), which added back to C(4,2)C(4, 2) yields R′(8,−6)R'(8, -6). Reflecting across y=−xy = -x exchanges the coordinates and negates them, giving R′′(6,−8)R''(6, -8). Rotating 90∘90^\circ counterclockwise about the origin maps (x,y)→(−y,x)(x, y) \rightarrow (-y, x), yielding R′′′(8,6)R'''(8, 6).
Estimated Time:3m 0s
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Topics

Transformations in the Coordinate Plane

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