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

Answer 1

To dilate a triangle with a magnitude of 3 in reference to the origin, you multiply each coordinate of the triangle's vertices by 3.

Given the original vertices:
- A(-2, -5)
- B(0, 3)
- C(6, -3)

Now we will apply the dilation:

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

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

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

Therefore, the new coordinates of the vertices after dilation will be:
- A′(-6, -15)
- B′(0, 9)
- C′(18, -9)

Thus the correct response is:
**A′(−6,−15), B′(0,9), and C′(18,−9)**.