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If you dilate triangle A (1,6) B (3,0) C (6,4) to form X'Y'Z' using a magnitude of 2 and a center of dilation at (2,0), what are the coordinates of the dilated triangle?



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Responses

A' (1.5,3), B' (2.5,0), C' (4,2)
A' (1.5,3), B' (2.5,0), C' (4,2)

A' (.5,3), B' (1.5,0), C' (3,2)
A' (.5,3), B' (1.5,0), C' (3,2)

A' (2,12), B' (6,0), C' (12,8)
A' (2,12), B' (6,0), C' (12,8)

A' (0,12), B' (4,0), C' (10,8)
A' (0,12), B' (4,0), C' (10,8)
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1 answer

To find the coordinates of the dilated triangle A'B'C' from triangle ABC using a center of dilation (2, 0) and a magnitude of 2, we can use the formula for dilation. The formula for a point \((x, y)\) dilated from a center \((x_c, y_c)\) with a scale factor \(k\) is given by:

\[ x' = x_c + k(x - x_c) \] \[ y' = y_c + k(y - y_c) \]

Given:

  • Center of dilation: \(C(2, 0)\)
  • Magnitude (scale factor): \(k = 2\)

Points:

  • A(1, 6)
  • B(3, 0)
  • C(6, 4)

Calculating the coordinates of A', B', and C':

  1. For A(1, 6): \[ x' = 2 + 2(1 - 2) = 2 + 2(-1) = 2 - 2 = 0 \] \[ y' = 0 + 2(6 - 0) = 0 + 12 = 12 \] So, \(A' = (0, 12)\).

  2. For B(3, 0): \[ x' = 2 + 2(3 - 2) = 2 + 2(1) = 2 + 2 = 4 \] \[ y' = 0 + 2(0 - 0) = 0 + 0 = 0 \] So, \(B' = (4, 0)\).

  3. For C(6, 4): \[ x' = 2 + 2(6 - 2) = 2 + 2(4) = 2 + 8 = 10 \] \[ y' = 0 + 2(4 - 0) = 0 + 8 = 8 \] So, \(C' = (10, 8)\).

Dilated Triangle Coordinates

Thus, the coordinates of the dilated triangle \(X'Y'Z'\) are:

  • \(A' (0, 12)\)
  • \(B' (4, 0)\)
  • \(C' (10, 8)\)

Correct Response

The answer matches the option: A' (0,12), B' (4,0), C' (10,8).