Use the image to answer the question.
A coordinate plane's x-axis ranges from negative 1 to 12 and its y-axis ranges from negative 5 to 6, both by 1-unit increments. Points upper A, upper B, upper C, and upper D are plotted forming a quadrilateral.
Translate quadrilateral ADCB 5 units to the left and 2 units down. What are the coordinates of new quadrilateral A′D′C′B′?
(1 point)
Responses
A′(−1,−4), D′(−1,1), C′(4,−2), B′(2,−5)
upper A prime left parenthesis negative 1 comma negative 4 right parenthesis , upper D prime left parenthesis negative 1 comma 1 right parenthesis , upper C prime left parenthesis 4 comma negative 2 right parenthesis , upper B prime left parenthesis 2 comma negative 5 right parenthesis
A′(9,0), D′(9,5), C′(13,2), B′(12,−1)
upper A prime left parenthesis 9 comma 0 right parenthesis , upper D prime left parenthesis 9 comma 5 right parenthesis , upper C prime left parenthesis 13 comma 2 right parenthesis , upper B prime left parenthesis 12 comma negative 1 right parenthesis
A′(−1,0), D′(−1,5), C′(4,2), B′(2,−1)
upper A prime left parenthesis negative 1 comma 0 right parenthesis , upper D prime left parenthesis negative 1 comma 5 right parenthesis , upper C prime left parenthesis 4 comma 2 right parenthesis , upper B prime left parenthesis 2 comma negative 1 right parenthesis
A′(9,−4), D′(9,1), C′(13,−2), B′(12,−5)
The correct answer is A′(−1,−4), D′(−1,1), C′(4,−2), B′(2,−5).
To translate the quadrilateral ADCB 5 units to the left and 2 units down, you need to subtract 5 from the x-coordinates and subtract 2 from the y-coordinates of each point.
Starting with the original coordinates:
A(-1, 6)
D(12, 6)
C(8, -5)
B(1, -5)
To translate 5 units to the left and 2 units down:
A' = (x-5, y-2)
D' = (x-5, y-2)
C' = (x-5, y-2)
B' = (x-5, y-2)
Applying the translation:
A' = (-1-5, 6-2) = (-6, 4)
D' = (12-5, 6-2) = (7, 4)
C' = (8-5, -5-2) = (3, -7)
B' = (1-5, -5-2) = (-4, -7)
Therefore, the coordinates of the new quadrilateral A'D'C'B' are:
A'(-6, 4), D'(7, 4), C'(3, -7), B'(-4, -7)