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¯¯¯¯¯¯¯¯
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 (−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 (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 .
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Answer 1

A′ = (1, 2) and B′ = (3, 3).

Reason: A dilation about the origin by scale factor 1/2 multiplies each coordinate by 1/2:
A′ = (2·1/2, 4·1/2) = (1,2), B′ = (6·1/2, 6·1/2) = (3,3).
Length AB = sqrt((6−2)^2 + (6−4)^2) = sqrt(20). Length A′B′ = sqrt((3−1)^2 + (3−2)^2) = sqrt(5) = (1/2)·sqrt(20), so A′B′ is half of AB.