Question

Use the image to answer the question.

Triangle upper A (-2,-5) upper B (0,3) upper C (6,-3) is graphed on a coordinate plane. The x-axis ranges from negative 6 to 6 in increments of 1. The y-axis ranges from negative 6 to 6 in increments of 1.

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′(−5,−8)
, B′(−3,0)
, and C′(3,0)
upper A prime left parenthesis negative 5 comma negative 8 right parenthesis , upper B prime left parenthesis negative 3 comma 0 right parenthesis , and upper C prime left parenthesis 3 comma 0 right parenthesis

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)

Answer 1

To dilate the triangle \( ABC \) with a magnitude of 3 in reference to the origin, we will multiply the coordinates of each vertex by 3.

The original coordinates are:
- \( A(-2, -5) \)
- \( B(0, 3) \)
- \( C(6, -3) \)

We perform the dilation as follows:

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

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

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

Thus, the coordinates of the new vertices \( A'B'C' \) after dilation are:
- \( A'(-6, -15) \)
- \( B'(0, 9) \)
- \( C'(18, -9) \)

This matches with the response:
**A′(−6,−15), B′(0,9), and C′(18,−9)**.