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)
To dilate a figure with a magnitude of 3 in reference to the origin, you multiply the x and y coordinates of each vertex by 3.
Using this information, the new coordinates of the vertices A', B', and C' can be found:
A' = (3 * x-coordinate of A, 3 * y-coordinate of A)
B' = (3 * x-coordinate of B, 3 * y-coordinate of B)
C' = (3 * x-coordinate of C, 3 * y-coordinate of C)
Therefore, the correct answer is:
A′(−5,−8), B′(−3,0), and C′(3,0)