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'?

Question

Coordinate A (-2,-5)
Coordinate B (0,3)
Coordinate C (6,-3)
A. A' (-2/3, -5/3), B'(0,1), and C'(2,-1)
B. A' (-6,-15), B' (0,9), and C' (18,-9)
C. A' (-5,-8), B' (-3,0), and C' (3,0)
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 certain magnitude, you multiply the coordinates of the point by the dilation factor. In this case, the dilation factor is 3.

Let's find the new coordinates for each vertex:

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

Now we compile the new coordinates for A', B', and C':

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

Thus, the new vertices after dilation are:

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

The answer matches option B: A' (-6, -15), B' (0, 9), and C' (18, -9).