Question

A (-3,3) -6 -4 < 6B (1,5) 4 2 x+ 10 (1 point) 10 -8 -2 -0- 0 -2- -4- C(4,-1) -6-D -8- (1,-5) < 4 -10- Paul is trying to dilate the figure in the image using a magnitude of 2 with reference to the origin. At what points would he draw the new vertices A'B'C'? Α΄(-32,32), Β΄(12,52), C'(2,-12) and D'(12,-52) A'(-6, 6), Β΄(2, 10), C'(8,-2), and D'(2, -10) A'(-5,1), Β΄(-1,3), C(2,-3) and D'(-1,-7) Ο Α΄(-1,5), Β΄(3,7), C'(6,1), and D'(3,-3)

Answer 1

To dilate a point \( (x, y) \) around the origin with a scale factor of \( k \), you can use the formula:

\[
(x', y') = (kx, ky)
\]

In this case, the scale factor \( k \) is 2.

Let's find the new vertices \( A', B', C', \) and \( D' \) using the given coordinates:

1. **Point A (-3, 3)**:
\[
A' = (2 \times -3, 2 \times 3) = (-6, 6)
\]

2. **Point B (1, 5)**:
\[
B' = (2 \times 1, 2 \times 5) = (2, 10)
\]

3. **Point C (4, -1)**:
\[
C' = (2 \times 4, 2 \times -1) = (8, -2)
\]

4. **Point D (1, -5)**:
\[
D' = (2 \times 1, 2 \times -5) = (2, -10)
\]

Now, we have the new vertices:

- \( A'(-6, 6) \)
- \( B'(2, 10) \)
- \( C'(8, -2) \)
- \( D'(2, -10) \)

Looking at the answer choices provided, the correct transformed vertices are:

**A'(-6, 6), B'(2, 10), C'(8, -2), and D'(2, -10)**.

So the correct answer is:
**A'(-6, 6), B'(2, 10), C'(8, -2), and D'(2, -10)**.