Answer:
[tex]T' = (-7,10)[/tex]
Step-by-step explanation:
Given
[tex]T = (10,7)[/tex]
[tex]r = 270^o[/tex] counterclockwise
Required
Graph of T'
The rule to this is:
[tex](x,y) \to (-y,x)[/tex]
So, we have:
[tex]T(10,7) \to T' (-7,10)[/tex]
Hence:
[tex]T' = (-7,10)[/tex]
See attachment for graph
