Question

a dilation of AB occurs using a scale factor of 1/2 and a center of dilation at the origin prove that A'B' is half the length of AB by finding points A'B' A' is at (4,8) and B' is at (12,12) A' is at (-1.-2) and B is at (-3,-3) A' is at (1,5) and B is at(3,6) or A' is at (1,2) and B' is at (3,3)

Answer 1

A dilation with center at the origin and scale factor k sends each point (x,y) to (kx, ky). For k = 1/2 this map is
A(x1,y1) -> A'(x1/2, y1/2) and B(x2,y2) -> B'(x2/2, y2/2).

General proof (distance formula):
AB = sqrt((x2 - x1)^2 + (y2 - y1)^2).
A'B' = sqrt((x2/2 - x1/2)^2 + (y2/2 - y1/2)^2)
= sqrt((1/4)((x2 - x1)^2 + (y2 - y1)^2))
= (1/2) sqrt((x2 - x1)^2 + (y2 - y1)^2)
= (1/2) AB.

So any dilation about the origin by 1/2 multiplies all segment lengths by 1/2.

Check the given coordinate examples (treating the listed points as A' and B', whose originals are twice those coordinates):

1) A' = (4,8), B' = (12,12). Originals: A=(8,16), B=(24,24).
AB = sqrt((24-8)^2+(24-16)^2)=sqrt(320). A'B' = sqrt((12-4)^2+(12-8)^2)=sqrt(80)=1/2 sqrt(320).

2) A' = (-1,-2), B' = (-3,-3). Originals: A=(-2,-4), B=(-6,-6).
AB = sqrt(20). A'B' = sqrt(5) = 1/2 sqrt(20).

3) A' = (1,5), B' = (3,6). Originals: A=(2,10), B=(6,12).
AB = sqrt(20). A'B' = sqrt(5) = 1/2 AB.

4) A' = (1,2), B' = (3,3). Originals: A=(2,4), B=(6,6).
AB = sqrt(20). A'B' = sqrt(5) = 1/2 AB.

In every case A'B' = (1/2) AB, as required.