Question

write the coordinates of the vertices after a dilation with a scale factor of \\(\frac{1}{3}\\), centered at the origin.

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of each vertex from the graph:

• \( A(-9, -9) \) (since it's at x = -9, y = -9)
• \( B(3, -9) \) (x = 3, y = -9)
• \( C(3, 6) \) (x = 3, y = 6)
• \( D(-9, 6) \) (x = -9, y = 6)

Step2: Apply dilation formula

The dilation formula centered at the origin with scale factor \( k \) is \( (x', y') = (k \cdot x, k \cdot y) \), where \( k = \frac{1}{3} \).

For \( A(-9, -9) \):

\( x' = \frac{1}{3} \cdot (-9) = -3 \)
\( y' = \frac{1}{3} \cdot (-9) = -3 \)
So \( A'(-3, -3) \)

For \( B(3, -9) \):

\( x' = \frac{1}{3} \cdot 3 = 1 \)
\( y' = \frac{1}{3} \cdot (-9) = -3 \)
So \( B'(1, -3) \)

For \( C(3, 6) \):

\( x' = \frac{1}{3} \cdot 3 = 1 \)
\( y' = \frac{1}{3} \cdot 6 = 2 \)
So \( C'(1, 2) \)

For \( D(-9, 6) \):

\( x' = \frac{1}{3} \cdot (-9) = -3 \)
\( y' = \frac{1}{3} \cdot 6 = 2 \)
So \( D'(-3, 2) \)

Answer:

\( A'(-3, -3) \), \( B'(1, -3) \), \( C'(1, 2) \), \( D'(-3, 2) \)