Question

Use the image to answer the question.

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

Answer 1

A′(−6,−15), B′(0,9), C′(18,−9)

Reason: dilation by 3 about the origin multiplies each vertex coordinate by 3. The original vertices are (−2,−5), (0,3), and (6,−3), so their images are 3× those coordinates above.