To find the coordinates of the dilated triangle \(X'Y'Z'\), we'll use the formula for dilation. The formula for dilating a point \((x, y)\) with a center of dilation \((x_c, y_c)\) and a scale factor \(k\) is given by:
\[ (x', y') = (x_c + k \cdot (x - x_c), y_c + k \cdot (y - y_c)) \]
In this case, the center of dilation is \(C(2, 0)\) and the scale factor is \(k = 2\). We'll apply this to each vertex of triangle \(ABC\).
Vertex A(1, 6)
\[ x_a' = 2 + 2 \cdot (1 - 2) = 2 + 2 \cdot (-1) = 2 - 2 = 0 \] \[ y_a' = 0 + 2 \cdot (6 - 0) = 0 + 2 \cdot 6 = 0 + 12 = 12 \] So, the new coordinates for \(X'\) are \((0, 12)\).
Vertex B(3, 0)
\[ x_b' = 2 + 2 \cdot (3 - 2) = 2 + 2 \cdot 1 = 2 + 2 = 4 \] \[ y_b' = 0 + 2 \cdot (0 - 0) = 0 + 2 \cdot 0 = 0 \] So, the new coordinates for \(Y'\) are \((4, 0)\).
Vertex C(6, 4)
\[ x_c' = 2 + 2 \cdot (6 - 2) = 2 + 2 \cdot 4 = 2 + 8 = 10 \] \[ y_c' = 0 + 2 \cdot (4 - 0) = 0 + 2 \cdot 4 = 0 + 8 = 8 \] So, the new coordinates for \(Z'\) are \((10, 8)\).
Final Result
The coordinates of the dilated triangle \(X'Y'Z'\) are:
- \(X' = (0, 12)\)
- \(Y' = (4, 0)\)
- \(Z' = (10, 8)\)
Thus, the coordinates of the dilated triangle are \((0, 12)\), \((4, 0)\), and \((10, 8)\).