To find the new vertices after dilating triangle XYZ with a center of dilation at point (4, 7) and a scale factor of \( \frac{1}{2} \), we can use the formula for dilation. The formula for finding the new point \( (x', y') \) after dilation is given by:
\[ x' = (1 - k) \cdot x_{0} + k \cdot x \] \[ y' = (1 - k) \cdot y_{0} + k \cdot y \]
where:
- \( (x, y) \) is the original point,
- \( (x_0, y_0) \) is the center of dilation,
- \( k \) is the scale factor.
In this case:
- \( k = \frac{1}{2} \)
- The center of dilation is \( (4, 7) \).
Let's calculate the new coordinates for each vertex:
-
Vertex X(0, 4): \[ x' = (1 - \frac{1}{2}) \cdot 4 + \frac{1}{2} \cdot 0 = \frac{1}{2} \cdot 4 + 0 = 2 \] \[ y' = (1 - \frac{1}{2}) \cdot 7 + \frac{1}{2} \cdot 4 = \frac{1}{2} \cdot 7 + 2 = 3.5 + 2 = 5.5 \] So, \( X' = (2, 5.5) \).
-
Vertex Y(2, 0): \[ x' = (1 - \frac{1}{2}) \cdot 4 + \frac{1}{2} \cdot 2 = \frac{1}{2} \cdot 4 + 1 = 2 + 1 = 3 \] \[ y' = (1 - \frac{1}{2}) \cdot 7 + \frac{1}{2} \cdot 0 = \frac{1}{2} \cdot 7 + 0 = 3.5 + 0 = 3.5 \] Thus, \( Y' = (3, 3.5) \).
-
Vertex Z(4, 7): \[ x' = (1 - \frac{1}{2}) \cdot 4 + \frac{1}{2} \cdot 4 = \frac{1}{2} \cdot 4 + 2 = 2 + 2 = 4 \] \[ y' = (1 - \frac{1}{2}) \cdot 7 + \frac{1}{2} \cdot 7 = \frac{1}{2} \cdot 7 + \frac{1}{2} \cdot 7 = 3.5 + 3.5 = 7 \] Thus, \( Z' = (4, 7) \).
So the coordinates of the new vertices of triangle X'Y'Z' are:
- \( X' = (2, 5.5) \)
- \( Y' = (3, 3.5) \)
- \( Z' = (4, 7) \)
In summary, the coordinates of the vertices of triangle X'Y'Z' after dilation are:
- \( X'(2, 5.5) \)
- \( Y'(3, 3.5) \)
- \( Z'(4, 7) \)