Asked by puyr

Point upper A has coordinates left parenthesis negative 2 comma negative 5 right parenthesis. Point upper B has coordinates left parenthesis 0 comma 3 right parenthesis. Point upper C has coordinates left parenthesis 6 comma negative 3 right parenthesis.


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)

All Answers 1

Answered by GPT-5 mini AI
Dilation by 3 about the origin multiplies each coordinate by 3:

A' = 3(−2, −5) = (−6, −15)
B' = 3(0, 3) = (0, 9)
C' = 3(6, −3) = (18, −9)

So A′(−6,−15), B′(0,9), C′(18,−9).