Question

A triangle A(-2,-5)B(0,3)C(6,-3) is drawn on a coordinate plane. dilate the figure with a magnitude of 3 in reference to the origin. what are the coordinates of the new vertices of A'B'C'?

Answer 1

To dilate a triangle (or any figure) from the origin with a scale factor (magnitude) of 3, you multiply the coordinates of each vertex by the scale factor.

The original vertices of the triangle are:
- A(-2, -5)
- B(0, 3)
- C(6, -3)

Now, applying the dilation with a scale factor of 3:

1. For vertex A:
\[
A' = (3 \times -2, 3 \times -5) = (-6, -15)
\]

2. For vertex B:
\[
B' = (3 \times 0, 3 \times 3) = (0, 9)
\]

3. For vertex C:
\[
C' = (3 \times 6, 3 \times -3) = (18, -9)
\]

Thus, the coordinates of the new vertices after the dilation are:
- A'(-6, -15)
- B'(0, 9)
- C'(18, -9)

So, the new vertices of triangle A'B'C' are:
- A'(-6, -15)
- B'(0, 9)
- C'(18, -9)