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).