Question

for the set of triangle vertices, find the coordinates of the vertices of the image after a dilation by the given scale factor and center of dilation. then grap and image on a separate sheet of paper.
a(9, 9), b(3, 3), c(6, 0), $k = \frac{1}{3}$, centered at b
$a$( , )
$b$( , )
$c$( , )

Explanation:

Step1: Recall the dilation formula

When dilating a point \( P(x, y) \) with center \( C(h, k) \) and scale factor \( s \), the coordinates of the image \( P'(x', y') \) are given by:

$$ x' = h + s(x - h) $$

$$ y' = k + s(y - k) $$

Step2: Dilate point \( A(9, 9) \) with center \( B(3, 3) \) and \( k=\frac{1}{3} \)

For \( x \)-coordinate of \( A' \):

$$ x' = 3 + \frac{1}{3}(9 - 3) = 3 + \frac{1}{3}(6) = 3 + 2 = 5 $$

For \( y \)-coordinate of \( A' \):

$$ y' = 3 + \frac{1}{3}(9 - 3) = 3 + \frac{1}{3}(6) = 3 + 2 = 5 $$

So, \( A'(5, 5) \)

Step3: Dilate point \( B(3, 3) \) with center \( B(3, 3) \) and \( k=\frac{1}{3} \)

For \( x \)-coordinate of \( B' \):

$$ x' = 3 + \frac{1}{3}(3 - 3) = 3 + 0 = 3 $$

For \( y \)-coordinate of \( B' \):

$$ y' = 3 + \frac{1}{3}(3 - 3) = 3 + 0 = 3 $$

So, \( B'(3, 3) \)

Step4: Dilate point \( C(6, 0) \) with center \( B(3, 3) \) and \( k=\frac{1}{3} \)

For \( x \)-coordinate of \( C' \):

$$ x' = 3 + \frac{1}{3}(6 - 3) = 3 + \frac{1}{3}(3) = 3 + 1 = 4 $$

For \( y \)-coordinate of \( C' \):

$$ y' = 3 + \frac{1}{3}(0 - 3) = 3 + \frac{1}{3}(-3) = 3 - 1 = 2 $$

So, \( C'(4, 2) \)

Answer:

\( A'(5, 5) \)
\( B'(3, 3) \)
\( C'(4, 2) \)