Answer:
[tex]y' = (6,4)[/tex]
Step-by-step explanation:
Given
[tex]y = (3,2)[/tex]
[tex]x = (4,0)[/tex]
[tex]z = (2,-2)[/tex]
[tex]D_{(O,2)}[/tex]
Required
The image of y
[tex]D_{(O,2)}[/tex] implies that y is dilated by a scale factor of 2 across the origin.
So, the image of y is:
[tex]y' = 2 * y[/tex]
[tex]y' = 2 * (3,2)[/tex]
[tex]y' = (6,4)[/tex]
