Question

what are the coordinates of the vertices of the image of triangle def after a dilation centered at f with a scale factor of \\(\frac{1}{3}\\) followed by a translation along the vector \\(\langle -3, 5 \
angle\\)?

Explanation:

First, we need to determine the coordinates of the original vertices \( D \), \( E \), and \( F \) from the graph. From the grid, we can see:

• \( D \) is at \( (2, 0) \)
• \( E \) is at \( (2, -3) \)
• \( F \) is at \( (-1, -3) \)

Step 1: Dilation centered at \( F \) with scale factor \( \frac{1}{3} \)

The formula for dilation centered at a point \( (h, k) \) with scale factor \( s \) is \( (x', y') = (h + s(x - h), k + s(y - k)) \)

For point \( D(2, 0) \):

\( h = -1 \), \( k = -3 \), \( s = \frac{1}{3} \)
\( x' = -1 + \frac{1}{3}(2 - (-1)) = -1 + \frac{1}{3}(3) = -1 + 1 = 0 \)
\( y' = -3 + \frac{1}{3}(0 - (-3)) = -3 + \frac{1}{3}(3) = -3 + 1 = -2 \)
So, \( D' \) is \( (0, -2) \)

For point \( E(2, -3) \):

\( x' = -1 + \frac{1}{3}(2 - (-1)) = -1 + \frac{1}{3}(3) = -1 + 1 = 0 \)
\( y' = -3 + \frac{1}{3}(-3 - (-3)) = -3 + \frac{1}{3}(0) = -3 \)
So, \( E' \) is \( (0, -3) \)

For point \( F(-1, -3) \):

\( x' = -1 + \frac{1}{3}(-1 - (-1)) = -1 + \frac{1}{3}(0) = -1 \)
\( y' = -3 + \frac{1}{3}(-3 - (-3)) = -3 + \frac{1}{3}(0) = -3 \)
So, \( F' \) is \( (-1, -3) \)

Step 2: Translation along vector \( \langle -3, 5

angle \)
The translation formula is \( (x'', y'') = (x' - 3, y' + 5) \)

For \( D'(0, -2) \):

\( x'' = 0 - 3 = -3 \)
\( y'' = -2 + 5 = 3 \)
So, \( D'' \) is \( (-3, 3) \)

For \( E'(0, -3) \):

\( x'' = 0 - 3 = -3 \)
\( y'' = -3 + 5 = 2 \)
So, \( E'' \) is \( (-3, 2) \)

For \( F'(-1, -3) \):

\( x'' = -1 - 3 = -4 \)
\( y'' = -3 + 5 = 2 \)
So, \( F'' \) is \( (-4, 2) \)

Answer:

\( D''(-3, 3) \), \( E''(-3, 2) \), \( F''(-4, 2) \)