Question

Graph the image of this figure after a dilation with a scale factor of 13 centered at the point (4, −2) .

Answer 1

Answer:

Just took the test and got it right!

Step-by-step explanation:

(look at the image down below :))

Answer 2

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) = $\frac{1}{3}$ and centered at point (4, -2).

The rule of dilation with $k= \frac{1}{3}$ and center at point (4,-2) is:

$(x, y) \rightarrow (\frac{1}{3}(x-4)+4 , \frac{1}{3}(y+2)-2)$

$(x, y) \rightarrow (\frac{1}{3}x-\frac{4}{3}+4 , \frac{1}{3}y+\frac{2}{3}-2)$

or

$(x, y) \rightarrow (\frac{1}{3}x+\frac{8}{3} , \frac{1}{3}y-\frac{4}{3})$

then, the dilation of the given figure are;

$A(1, 10) \rightarrow (\frac{1}{3}\cdot 1+\frac{8}{3} , \frac{1}{3} \cdot 10-\frac{4}{3})$ = $(\frac{1}{3}+\frac{8}{3} , \frac{10}{3}-\frac{4}{3})$ =$(\frac{9}{3} , \frac{6}{3})$ = A'(3 , 2)

$B(-2, 4) \rightarrow (\frac{1}{3}\cdot -2+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})$ =$(-\frac{2}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})$ =$(\frac{6}{3} , \frac{0}{3})$ =B'(2 , 0)

$C(7, 4) \rightarrow (\frac{1}{3}\cdot 7+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})$ =$(\frac{7}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})$ =$(\frac{15}{3} , \frac{0}{3})$ = 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: