Question

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′ ? (1 point) Responses A′(1,−2) , B′(3,6) , and C′(9,0) upper A prime left parenthesis 1 comma negative 2 right parenthesis , upper B prime left parenthesis 3 comma 6 right parenthesis , and upper C prime left parenthesis 9 comma 0 right parenthesis A′(−6,−15) , B′(0,9) , and C′(18,−9) upper A prime left parenthesis negative 6 comma negative 15 right parenthesis , upper B prime left parenthesis 0 comma 9 right parenthesis , and upper C prime left parenthesis 18 comma negative 9 right parenthesis A′(−23,−53) , B′(0,1) , and C′(2,−1) upper A prime left parenthesis negative Start Fraction 2 over 3 End Fraction comma negative Start Fraction 5 over 3 End Fraction right parenthesis , upper B prime left parenthesis 0 comma 1 right parenthesis , and upper C prime left parenthesis 2 comma negative 1 right parenthesis A′(−5,−8) , B′(−3,0) , and C′(3,0)
Corner B: (0, 3)
Corner A: (-2, -5)
Corner C: (6, -3)

Answer 1

To dilate a triangle around the origin with a magnitude of 3, you multiply each coordinate of the vertices by 3.

Given the coordinates of the triangle:

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

We calculate the new coordinates as follows:

- A' = (3 * -2, 3 * -5) = (-6, -15)
- B' = (3 * 0, 3 * 3) = (0, 9)
- C' = (3 * 6, 3 * -3) = (18, -9)

Thus, the coordinates of the new vertices A′, B′, and C′ are:

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

The correct response is:

**A′(−6,−15) , B′(0,9) , and C′(18,−9)**