Answer :
Answer:
The coordinates of the vertices of [tex]\bigtriangleup A'B'C'[/tex] are [tex]A'=(-2,8)[/tex], [tex]B'=(-17,-4)[/tex] and [tex]C'=(-9,0)[/tex].
Step-by-step explanation:
Let be [tex]A =(7,3)[/tex], [tex]B = (-8,6)[/tex] and [tex]C = (0,-5)[/tex] the vertices of the original triangle, as well as [tex]\vec t = \left<-9, 5\right>[/tex], the coordinates of the vertices of the new triangle are, respectively:
[tex]A' = A + \vec t[/tex]
[tex]A' = (7,3)+(-9,5)[/tex]
[tex]A' = (7-9,3+5)[/tex]
[tex]A'=(-2,8)[/tex]
[tex]B' = B + \vec t[/tex]
[tex]B' = (-8,6)+(-9,5)[/tex]
[tex]B' = (-8-9,-9+5)[/tex]
[tex]B'=(-17,-4)[/tex]
[tex]C' = C + \vec t[/tex]
[tex]C' = (0,-5)+(-9,5)[/tex]
[tex]C' = (0-9,-5+5)[/tex]
[tex]C'=(-9,0)[/tex]
The coordinates of the vertices of [tex]\bigtriangleup A'B'C'[/tex] are [tex]A'=(-2,8)[/tex], [tex]B'=(-17,-4)[/tex] and [tex]C'=(-9,0)[/tex].