Question

If you dilate triangle XY Z to form X'Y'Z' using a magnitude of 2 and a center of dilation at (2, 0), what are the coordinates of the dilated triangle? X (2,4) Z (7,2) Y (6,0)
(1 point)
O X'(6,4), Y'(10, 0), Z' (11, 4)
O X'(2,2), Y'(4,0), Z'(4.5, 1)
• X'(2,6), Y'(8, 0), Z' (9.5, 3)
O X'(2,8), Y'(10, 0), Z'(12, 4)

Answer 1

To find the coordinates of the dilated triangle \( X'Y'Z' \) after dilating triangle \( XYZ \) using a magnitude of 2 and a center of dilation at (2, 0), we'll use the dilation formula. The formula for dilation from a center point \( (a, b) \) is given by:

\[
(x', y') = (a + k(x - a), b + k(y - b))
\]

where \( k \) is the dilation factor.

In this case, the center of dilation is \( (2, 0) \) and the dilation factor is \( k = 2 \).

Let's calculate the coordinates of the dilated points:

1. **For point \( X (2, 4) \)**:
\[
X' = (2 + 2(2 - 2), 0 + 2(4 - 0)) = (2 + 0, 0 + 8) = (2, 8)
\]

2. **For point \( Y (6, 0) \)**:
\[
Y' = (2 + 2(6 - 2), 0 + 2(0 - 0)) = (2 + 8, 0 + 0) = (10, 0)
\]

3. **For point \( Z (7, 2) \)**:
\[
Z' = (2 + 2(7 - 2), 0 + 2(2 - 0)) = (2 + 10, 0 + 4) = (12, 4)
\]

Now we have the coordinates for the dilated points:
- \( X' = (2, 8) \)
- \( Y' = (10, 0) \)
- \( Z' = (12, 4) \)

Therefore, the coordinates of the dilated triangle \( X'Y'Z' \) are \( X'(2, 8), Y'(10, 0), Z'(12, 4) \).

The correct answer is:

**X'(2,8), Y'(10, 0), Z'(12, 4)**