Question

Triangle XYZ is drawn on a coordinate plane with vertices X(0,4) , Y(2,0) , and Z(4,7) . If you dilate the triangle to form triangle X′Y′Z′ using a magnitude of 1/2 and a center of dilation at (4,7) , what are the coordinate points of the new vertices?

A. X′(−4,1) , Y′(0,−7) , Z′(4,7)
B. X′(−4,−3) , Y′(−2,−7) , Z′(0,0)
C. X′(4,11) , Y′(6,7) , Z′(8,14)
D. X′(2,5.5) , Y′(3,3.5) , Z′(4,7)

Answer 1

To find the new coordinates of the vertices \( X' \), \( Y' \), and \( Z' \) after dilating triangle \( XYZ \) with a center of dilation at \( (4, 7) \) and a magnitude of \( \frac{1}{2} \), we use the formula for dilation:

\[
X' = (1 - k)C + kX
\]

where \( k \) is the dilation factor, \( C \) is the center of dilation, and \( X \) is the original point.

Given:
- Center of dilation \( C = (4, 7) \)
- Dilation factor \( k = \frac{1}{2} \)

### Step 1: Calculate the new coordinates for vertex \( X(0, 4) \)

\[
X' = \left( 4 + \frac{1}{2}(0 - 4), 7 + \frac{1}{2}(4 - 7) \right)
\]
\[
X' = \left( 4 - 2, 7 - \frac{3}{2} \right)
\]
\[
X' = (2, 5.5)
\]

### Step 2: Calculate the new coordinates for vertex \( Y(2, 0) \)

\[
Y' = \left( 4 + \frac{1}{2}(2 - 4), 7 + \frac{1}{2}(0 - 7) \right)
\]
\[
Y' = \left( 4 - 1, 7 - \frac{7}{2} \right)
\]
\[
Y' = (3, 3.5)
\]

### Step 3: Calculate the new coordinates for vertex \( Z(4, 7) \)

Since \( Z \) is the center of dilation:
\[
Z' = (4, 7) \quad \text{(remains the same)}
\]

### Final Result

Thus, the coordinates of the new vertices \( X' \), \( Y' \), and \( Z' \) after dilation are:
\[
X' = (2, 5.5), \quad Y' = (3, 3.5), \quad Z' = (4, 7)
\]

The correct answer is:

**D. \( X'(2, 5.5), Y'(3, 3.5), Z'(4, 7) \)**.