Question

The coordinates of points upper A and upper B are located as follows: upper A at left parenthesis 2 comma 4 right parenthesis and upper B at left parenthesis 6 comma 6 right parenthesis. The third unlabeled point is located at the origin. The line connecting point upper A with upper B is solid. The line connecting upper B with the origin point is a hybrid solid and dotted line that extends past the origin and past upper B. The line connecting upper A with the origin point is a dotted line that extends past the origin and past upper A.

A dilation of AB¯¯¯¯¯¯¯¯
𝐴
𝐵
¯
occurs using a scale factor of 12
1
2
and a center of dilation at the origin. Prove that A′B′¯¯¯¯¯¯¯¯¯¯
𝐴
′
𝐵
′
¯
is half the length of AB¯¯¯¯¯¯¯¯
𝐴
𝐵
¯
by finding the points A′
𝐴
′
and B′
𝐵
′
.

(1 point)
Responses

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

Answer 1

A dilation about the origin with scale factor k = 1/2 sends (x,y) to (kx, ky). Thus

- A' = (1/2·2, 1/2·4) = (1, 2).
- B' = (1/2·6, 1/2·6) = (3, 3).

Check lengths:
- AB = sqrt((6−2)^2 + (6−4)^2) = sqrt(16+4) = sqrt(20) = 2·sqrt(5).
- A'B' = sqrt((3−1)^2 + (3−2)^2) = sqrt(4+1) = sqrt(5).

So A'B' = (1/2)·AB, as required.