Question

y
6
5
4
B (0,3)
3
210
x
6 -5 -4 -3
-2
-1 0
1
2
3
4
5
6
-1+
-2
-3
C (6 ,- 3)
-4
-5
A (-2,5)
-6-
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′(−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)
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′(−6,−15)
, B′(0,9)
, and C′(18,−9)

Answer 1

To find the coordinates of the new vertices \( A'B'C' \) after dilating the triangle with a magnitude of 3 in reference to the origin, you need to multiply the coordinates of each vertex by the dilation factor.

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

Now, we will perform the dilation:

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

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

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

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

Therefore, the response corresponding to your problem is:

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