Answer:
C) (-2, 4), (6,8) is the correct answer.
Step-by-step explanation:
Given that line segment AB:
A (-3, 6) and B (9, 12) is dilated with a scale factor 2/3 about the origin.
First of all, let us calculate the distance AB using the distance formula:
[tex]D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Here,
[tex]x_2=9\\x_1=-3\\y_2=12\\y_1=6[/tex]
Putting all the values and finding AB:
[tex]AB = \sqrt{(9-(-3))^2+(12-6)^2}\\\Rightarrow AB = \sqrt{(12)^2+(6)^2}\\\Rightarrow AB = \sqrt{144+36}\\\Rightarrow AB = \sqrt{180}\\\Rightarrow AB = 6\sqrt{5}\ units[/tex]
It is given that AB is dilated with a scale factor of [tex]\frac{2}{3}[/tex].
[tex]x_2'=\dfrac{2}{3}\times x_2=\dfrac{2}{3}\times9=6\\x_1'=\dfrac{2}{3}\times x_1=\dfrac{2}{3}\times-3=-2\\y_2'=\dfrac{2}{3}\times y_2=\dfrac{2}{3}\times 12=8\\y_1'=\dfrac{2}{3}\times y_1=\dfrac{2}{3}\times 6=4[/tex]
So, the new coordinates are A'(-2,4) and B'(6,8).
Verifying this by calculating the distance A'B':
[tex]A'B' = \sqrt{(6-(-2))^2+(8-4)^2}\\\Rightarrow A'B' = \sqrt{(8)^2+(4)^2}\\\Rightarrow A'B' = \sqrt{64+16}\\\Rightarrow A'B' = \sqrt{80}\\\Rightarrow A'B' = 4\sqrt{5}\ units = \dfrac{2}{3}\times AB[/tex]
So, option C) (-2, 4), (6,8) is the correct answer.
