Question

A (-2,-5)
B (0,3)
C (6,-3)

A triangle 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'?

a. A' (-5,-8), B' (-3,0), and C' (3,0)
b. A' (- 2/5,- 5/3), B' (0,1), and C' (2,-1)
c. A' (-6,-15), B' (0,9), and C' (18,-9)
d. A' (1,-2), B' (3,6), and C' (9,0)

Answer 1

To dilate a point in the coordinate plane with respect to the origin by a magnitude of \( k \), you multiply both the x and y coordinates by \( k \).

For a dilation with a magnitude of \( 3 \), the transformation can be described as follows:

If \( P(x, y) \) is a point, then the dilated point \( P'(x', y') \) will be:

\[
x' = k \cdot x
\]
\[
y' = k \cdot y
\]

Applying this to the vertices of triangle ABC:

1. For point A (-2, -5):

\[
A' = (3 \cdot -2, 3 \cdot -5) = (-6, -15)
\]

2. For point B (0, 3):

\[
B' = (3 \cdot 0, 3 \cdot 3) = (0, 9)
\]

3. For point C (6, -3):

\[
C' = (3 \cdot 6, 3 \cdot -3) = (18, -9)
\]

So the coordinates of the new vertices after dilation would be:

- A' (-6, -15)
- B' (0, 9)
- C' (18, -9)

This corresponds to option c. Therefore, the correct answer is:

**c. A' (-6, -15), B' (0, 9), and C' (18, -9)**