Question

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 negative 1 comma negative 2 right parenthesis and upper B prime is at left parenthesis negative 3 comma negative 3 right parenthesis .

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 (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 (1,5)
and B′
is at (3,6)
.

Answer 1

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

Reason: The dilation center is the origin with scale factor 1/2, so (x,y) → (x/2,y/2). The original points are A(2,4) and B(6,6), so A′=(2/2,4/2)=(1,2) and B′=(6/2,6/2)=(3,3).

Lengths: AB = sqrt((6−2)^2+(6−4)^2)=sqrt(16+4)=sqrt20, A′B′ = sqrt((3−1)^2+(3−2)^2)=sqrt(4+1)=sqrt5 = (1/2)·sqrt20, so A′B′ is half of AB.