To find the coordinates of the dilated triangle X'Y'Z', we need to apply the dilation transformation on each vertex of triangle ABC. The formula for dilation with a center (x_c, y_c) and a scale factor \( k \) is given by:
\[ (x', y') = (x_c + k(x - x_c), y_c + k(y - y_c)) \]
Given your triangle vertices:
- \( A(1, 6) \)
- \( B(3, 0) \)
- \( C(6, 4) \)
And the center of dilation at \( (2, 0) \) with a magnitude (scale factor) of \( 2 \), we can calculate the new coordinates for each vertex.
1. For vertex A(1, 6): \[ x' = 2 + 2(1 - 2) = 2 + 2(-1) = 2 - 2 = 0 \] \[ y' = 0 + 2(6 - 0) = 0 + 2(6) = 0 + 12 = 12 \] So, \( A' = (0, 12) \).
2. For vertex B(3, 0): \[ x' = 2 + 2(3 - 2) = 2 + 2(1) = 2 + 2 = 4 \] \[ y' = 0 + 2(0 - 0) = 0 + 2(0) = 0 + 0 = 0 \] So, \( B' = (4, 0) \).
3. For vertex C(6, 4): \[ x' = 2 + 2(6 - 2) = 2 + 2(4) = 2 + 8 = 10 \] \[ y' = 0 + 2(4 - 0) = 0 + 2(4) = 0 + 8 = 8 \] So, \( C' = (10, 8) \).
Putting it all together, the coordinates of the dilated triangle \( X'Y'Z' \) are:
- \( A' = (0, 12) \)
- \( B' = (4, 0) \)
- \( C' = (10, 8) \)
Therefore, the correct response is: A' (0,12), B' (4,0), C' (10,8).
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