Question

Triangle ABC has the following coordinates: A=(5,-5), B=(3,-3), C=(5,-3) What are the coordinates of triangle A'B'C' if it is created by dilating triangle ABC with the origin (0,0) as the center of dilation and with a scale factor of 3?

Answer 1

Answer:A' = (15, -15), B' = (9, -9), and C' = (15, -9)

Explanation:

To dilate triangle ABC with a center of dilation at the origin (0,0) and a scale factor of 3, you need to multiply the coordinates of each vertex by the scale factor.

Let's calculate the coordinates of triangle A'B'C':

For point A:

x-coordinate of A' = scale factor * x-coordinate of A = 3 * 5 = 15

y-coordinate of A' = scale factor * y-coordinate of A = 3 * (-5) = -15

Therefore, A' = (15, -15)

For point B:

x-coordinate of B' = scale factor * x-coordinate of B = 3 * 3 = 9

y-coordinate of B' = scale factor * y-coordinate of B = 3 * (-3) = -9

Therefore, B' = (9, -9)

For point C:

x-coordinate of C' = scale factor * x-coordinate of C = 3 * 5 = 15

y-coordinate of C' = scale factor * y-coordinate of C = 3 * (-3) = -9

Therefore, C' = (15, -9)

Hence, the correct coordinates of triangle A'B'C' are A' = (15, -15), B' = (9, -9), and C' = (15, -9).