Asked by jime

Use the image to answer the question.

A coordinate plane's x-axis ranges from negative 2 to 8 and its y-axis ranges from negative 1 to 7, both by 1-unit increments. Point upper A, point upper B, and an unlabeled third point are plotted and connected by lines.

A dilation of AB¯¯¯¯¯¯¯¯
occurs using a scale factor of 12
and a center of dilation at the origin. Prove that A′B′¯¯¯¯¯¯¯¯¯¯
is half the length of AB¯¯¯¯¯¯¯¯
by finding the points A′
and B′
.

(1 point)
Responses

A′
is at (1,2)
and B′
is at (3,3)
.
upper A prime is at left parenthesis 1 comma 2 right parenthesis and upper B prime is at left parenthesis 3 comma 3 right parenthesis .

A′
is at (1,5)
and B′
is at (3,6)
.
upper A prime is at left parenthesis 1 comma 5 right parenthesis and upper B prime is at left parenthesis 3 comma 6 right parenthesis .

A′
is at (4,8)
and B′
is at (12,12)
.
upper A prime is at left parenthesis 4 comma 8 right parenthesis and upper B prime is at left parenthesis 12 comma 12 right parenthesis .

A′
is at (−1,−2)
and B′
is at (−3,−3)
.

Answers

There are no human answers yet.
Answered by GPT-5 mini
The center is the origin and the scale factor is 1/2, so (x,y) → (x/2,y/2). From the graph A = (2,4) and B = (6,6).

Thus A′ = (2/2, 4/2) = (1,2) and B′ = (6/2, 6/2) = (3,3).

Check lengths: AB = sqrt((6−2)^2+(6−4)^2) = sqrt(16+4) = 2√5, and A′B′ = sqrt((3−1)^2+(3−2)^2) = √5, which is half of AB.

So the correct choice is A′ = (1,2) and B′ = (3,3).