To find the coordinates of the new vertices of triangle \(X'Y'Z'\) after dilating triangle \(XYZ\) with a center of dilation at \((4,7)\) and a magnitude of 12, we can use the formula for dilation:
\[ X' = (kx + (1-k)h, ky + (1-k)k) \]
where:
- \((x, y)\) are the original coordinates of the vertex,
- \((h, k)\) is the center of dilation,
- \(k\) is the scale factor.
Since we are using a magnitude (scale factor) of 12, \(k = 12\).
Let's calculate the new coordinates of each vertex using the center of dilation at \((4, 7)\):
-
For vertex \(X(0, 4)\): \[ X' = (12 * (0 - 4) + 4, 12 * (4 - 7) + 7) \] \[ = (12*(-4) + 4, 12*(-3) + 7) = (-48 + 4, -36 + 7) = (-44, -29) \]
-
For vertex \(Y(2, 0)\): \[ Y' = (12 * (2 - 4) + 4, 12 * (0 - 7) + 7) \] \[ = (12*(-2) + 4, 12*(-7) + 7) = (-24 + 4, -84 + 7) = (-20, -77) \]
-
For vertex \(Z(4, 7)\): \[ Z' = (12 * (4 - 4) + 4, 12 * (7 - 7) + 7) \] \[ = (12*(0) + 4, 12*(0) + 7) = (4, 7) \]
After calculating, the new coordinates of the vertices are:
- \(X'(-44, -29)\)
- \(Y'(-20, -77)\)
- \(Z'(4, 7)\) (remains the same since it's the center of dilation)
From the answer choices, none of these coordinates match the responses given. Could you please double-check the scale factor or the center of dilation?