To dilate triangle ABC with a center of dilation at (2, 0) and a magnitude of 2, we need to apply the formula for dilation. The formula for the coordinates of the dilated points is given by:
\[ X' = (1 + k)(X - C_x) + C_x \] \[ Y' = (1 + k)(Y - C_y) + C_y \]
where \( (C_x, C_y) \) is the center of dilation, \( (X, Y) \) is the original point, and \( k \) is the scale factor. Here, the center of dilation is (2, 0), and the magnitude (scale factor) \( k = 2 - 1 = 1 \).
Thus, using the formula, for a point \( (X, Y) \):
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First, calculate the new X coordinate: \[ X' = (1 + 1)(X - 2) + 2 = 2(X - 2) + 2 = 2X - 4 + 2 = 2X - 2 \]
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Next, calculate the new Y coordinate: \[ Y' = (1 + 1)(Y - 0) + 0 = 2Y \]
Now, to determine the coordinates of the dilated triangle, we need the original coordinates of points A, B, and C. However, those coordinates are not provided in your question, so we cannot directly calculate the new coordinates of A', B', and C'.
Instead, if we illustrate the transformation with hypothetical points, let's say for example \( A(1, 3) \), \( B(2, 0) \), and \( C(3, 2) \), applying the dilation would yield:
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For point \( A(1, 3) \): \[ A'(X) = 2(1) - 2 = 0 \] \[ A'(Y) = 2(3) = 6 \] gives \( A'(0, 6) \).
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For point \( B(2, 0) \): \[ B'(X) = 2(2) - 2 = 2 \] \[ B'(Y) = 2(0) = 0 \] gives \( B'(2, 0) \).
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For point \( C(3, 2) \): \[ C'(X) = 2(3) - 2 = 4 \] \[ C'(Y) = 2(2) = 4 \] gives \( C'(4, 4) \).
Therefore, depending on the original points you have for \( A, B, C \), you can use the formula above to find the coordinates of the dilated triangle.
Please provide the coordinates of triangle ABC for a specific answer or confirm the original coordinates that have been given!
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