Question

X(2,4), Z(7,2,), Y(6,0) if you dilated triangle XYZ 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?

Answer 1

To dilate the triangle \( XYZ \) with a center of dilation at \( (2,0) \) and a magnitude of \( 2 \), we can use the following formula for dilation:

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

Where:
- \( X \) are the original coordinates,
- \( C \) is the center of dilation,
- \( k \) is the magnitude of the dilation minus 1 (for enlargement, \( k > 0 \)).

Given:
- \( C = (2, 0) \) (the center of dilation)
- \( k = 2 \) (the dilation factor)

So, for each vertex of the triangle, we will apply the formula.

**1. For point \( X(2,4) \):**

\[
X' = (1 + 2)(X - C) + C
\]
\[
= 3 \cdot [(2,4) - (2,0)] + (2,0)
\]
\[
= 3 \cdot (0, 4) + (2,0) = (0, 12) + (2, 0) = (2, 12)
\]

**2. For point \( Y(6,0) \):**

\[
Y' = (1 + 2)(Y - C) + C
\]
\[
= 3 \cdot [(6,0) - (2,0)] + (2,0)
\]
\[
= 3 \cdot (4, 0) + (2,0) = (12, 0) + (2, 0) = (14, 0)
\]

**3. For point \( Z(7,2) \):**

\[
Z' = (1 + 2)(Z - C) + C
\]
\[
= 3 \cdot [(7,2) - (2,0)] + (2,0)
\]
\[
= 3 \cdot (5, 2) + (2,0) = (15, 6) + (2, 0) = (17, 6)
\]

Putting it all together, the coordinates of the dilated triangle \( X'Y'Z' \) are:
- \( X'(2, 12) \)
- \( Y'(14, 0) \)
- \( Z'(17, 6) \)

Thus, the final coordinates of the dilated triangle \( X'Y'Z' \) are \( (2, 12), (14, 0), (17, 6) \).