Answer:
From the given triangle figure;
Labelled the triangle as A , B and C.
The coordinates of this triangle ABC are;
A = (1, 10) ,
B = (-2, 4)
C = (7, 4).
Given : Scale factor(k) = [tex]\frac{1}{3}[/tex] and centered at point (4, -2).
The rule of dilation with [tex]k= \frac{1}{3}[/tex] and center at point (4,-2) is:
[tex](x, y) \rightarrow (\frac{1}{3}(x-4)+4 , \frac{1}{3}(y+2)-2)[/tex]
[tex](x, y) \rightarrow (\frac{1}{3}x-\frac{4}{3}+4 , \frac{1}{3}y+\frac{2}{3}-2)[/tex]
or
[tex](x, y) \rightarrow (\frac{1}{3}x+\frac{8}{3} , \frac{1}{3}y-\frac{4}{3})[/tex]
then, the dilation of the given figure are;
[tex]A(1, 10) \rightarrow (\frac{1}{3}\cdot 1+\frac{8}{3} , \frac{1}{3} \cdot 10-\frac{4}{3})[/tex] = [tex](\frac{1}{3}+\frac{8}{3} , \frac{10}{3}-\frac{4}{3})[/tex] =[tex] (\frac{9}{3} , \frac{6}{3})[/tex] = A'(3 , 2)
[tex]B(-2, 4) \rightarrow (\frac{1}{3}\cdot -2+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})[/tex] =[tex] (-\frac{2}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})[/tex] =[tex] (\frac{6}{3} , \frac{0}{3})[/tex] =B'(2 , 0)
[tex]C(7, 4) \rightarrow (\frac{1}{3}\cdot 7+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})[/tex] =[tex] (\frac{7}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})[/tex] =[tex] (\frac{15}{3} , \frac{0}{3})[/tex] = C'(5 , 0)
The coordinates of dilation images are:
A' = (3,2) , B' = (2, 0) and C' = (5, 0)
You can see the graph of the dilated image as shown below:
