Question

Triangle XYZ is drawn on a coordinate plane with vertices X(0,4)Y(2,0)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

Answer 1

To find the coordinates of the new vertices after dilating triangle XYZ with a center at point \( C(4, 7) \) and a magnitude of \( \frac{1}{2} \), you can use the formula for dilation:

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

where \( k \) is the scale factor (in this case, \( k = \frac{1}{2} \)), \( X \) is the original point, and \( C \) is the center of dilation.

The original vertices are:
- \( X(0, 4) \)
- \( Y(2, 0) \)
- \( Z(4, 7) \)

**Calculating the new coordinates:**

1. **For vertex \( X(0, 4) \):**

\[
X' = (4, 7) + \frac{1}{2}((0, 4) - (4, 7))
\]
\[
= (4, 7) + \frac{1}{2}((-4, -3))
\]
\[
= (4, 7) + (-2, -1.5)
\]
\[
= (4 - 2, 7 - 1.5) = (2, 5.5)
\]

2. **For vertex \( Y(2, 0) \):**

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

3. **For vertex \( Z(4, 7) \):**

Since \( Z \) is the center of dilation, its coordinates will remain the same:

\[
Z' = Z(4, 7)
\]

So, after the dilation, the coordinates of triangle \( X'Y'Z' \) are:
- \( X'(2, 5.5) \)
- \( Y'(3, 3.5) \)
- \( Z'(4, 7) \)

In summary, the new vertices after dilation are:
- \( X'(2, 5.5) \)
- \( Y'(3, 3.5) \)
- \( Z'(4, 7) \)