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.
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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′(−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′(1,−2)
, B′(3,6)
, and C′(9,0)

Answer 1

Dilation by 3 about the origin multiplies each coordinate by 3. The original vertices are (-2,-5), (0,3), (6,-3), so

A′ = 3(-2,-5) = (-6,-15), B′ = 3(0,3) = (0,9), C′ = 3(6,-3) = (18,-9).