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Question


If you dilate triangle ABC 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?

A is (1,6)
B is (3,0)
C is (6,4)

(1 point)
Responses

A' (1.5,3), B' (2.5,0), C' (4,2)
A' (1.5,3), B' (2.5,0), C' (4,2)

A' (.5,3), B' (1.5,0), C' (3,2)
A' (.5,3), B' (1.5,0), C' (3,2)

A' (2,12), B' (6,0), C' (12,8)
A' (2,12), B' (6,0), C' (12,8)

A' (0,12), B' (4,0), C' (10,8)
A' (0,12), B' (4,0), C' (10,8)
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Answer 1

To find the coordinates of the dilated triangle \( A'B'C' \) after dilating triangle \( ABC \) using a magnitude of 2 and a center of dilation at \( (2, 0) \), we can apply the following formula for dilation:

\[
(x', y') = (k(x - c_x) + c_x, k(y - c_y) + c_y)
\]

where:
- \( (x, y) \) are the original coordinates,
- \( (c_x, c_y) \) is the center of dilation,
- \( k \) is the magnitude of dilation.

Given:
- Center of dilation \( (c_x, c_y) = (2, 0) \)
- Magnitude of dilation \( k = 2 \)

Now we will calculate the coordinates of points \( A' \), \( B' \), and \( C' \).

1. **For point A (1, 6)**:
\[
A' = (2 + 2 \cdot (1 - 2), 0 + 2 \cdot (6 - 0)) = (2 + 2 \cdot (-1), 0 + 12) = (0, 12)
\]

2. **For point B (3, 0)**:
\[
B' = (2 + 2 \cdot (3 - 2), 0 + 2 \cdot (0 - 0)) = (2 + 2 \cdot (1), 0 + 0) = (4, 0)
\]

3. **For point C (6, 4)**:
\[
C' = (2 + 2 \cdot (6 - 2), 0 + 2 \cdot (4 - 0)) = (2 + 2 \cdot (4), 0 + 8) = (2 + 8, 8) = (10, 8)
\]

Thus, the coordinates of the dilated triangle \( A'B'C' \) are:
- \( A' (0, 12) \)
- \( B' (4, 0) \)
- \( C' (10, 8) \)

The correct response from the options you provided is:
**A' (0, 12), B' (4, 0), C' (10, 8)**.