To find the coordinates of the dilated triangle X'Y'Z' when triangle XYZ is dilated with a magnitude of 2 and a center of dilation at (2,0), we can use the dilation formula:
\[ (X', Y') = \left( x_c + k(x - x_c), y_c + k(y - y_c) \right) \]
where \((x_c, y_c)\) is the center of dilation, \(k\) is the magnitude of the dilation, and \((x, y)\) are the original coordinates of the point.
Here, the center of dilation is \((2,0)\) and the magnitude of dilation \(k\) is \(2\).
We will calculate the new coordinates for each vertex of triangle XYZ:
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For point X (2, 4): \[ X' = \left( 2 + 2(2 - 2), 0 + 2(4 - 0) \right) = \left( 2 + 0, 0 + 8 \right) = (2, 8) \]
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For point Y (6, 0): \[ Y' = \left( 2 + 2(6 - 2), 0 + 2(0 - 0) \right) = \left( 2 + 2(4), 0 + 0 \right) = (2 + 8, 0) = (10, 0) \]
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For point Z (7, 2): \[ Z' = \left( 2 + 2(7 - 2), 0 + 2(2 - 0) \right) = \left( 2 + 2(5), 0 + 2(2) \right) = (2 + 10, 0 + 4) = (12, 4) \]
Now we have the coordinates of the dilated triangle X'Y'Z':
- X' = (2, 8)
- Y' = (10, 0)
- Z' = (12, 4)
Thus, the coordinates of the dilated triangle X'Y'Z' are:
- \(X'(2, 8)\)
- \(Y'(10, 0)\)
- \(Z'(12, 4)\)
Comparing these coordinates with the provided options, the correct answer is:
D. X'(2,8), Y'(10,0), Z'(12,4).
1 answer