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′(−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′(−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′(−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
oringal choords
A -2, -5
B 0, 3
C 6, -3

Answer 1

To dilate the triangle with vertices A(-2, -5), B(0, 3), and C(6, -3) with a magnitude of 3 in reference to the origin, we apply the dilation formula for each vertex:

\[
(x', y') = (k \cdot x, k \cdot y)
\]

where \(k\) is the dilation factor (in this case, \(k = 3\)) and \((x, y)\) are the original coordinates.

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

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

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

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

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