Question

write the coordinates of the vertices after a dilation with a scale factor of \\(\frac{1}{4}\\), centered at the origin.
\\(r(\square, \square)\\)
\\(s(\square, \square)\\)
\\(t(\square, \square)\\)
\\(u(\square, \square)\\)

Explanation:

Step1: Find original coordinates

From the graph, we identify the original coordinates:

• \( R(-8, -8) \)
• \( S(-8, 8) \)
• \( T(4, 8) \)
• \( U(4, -8) \)

Step2: Apply dilation rule

The rule for dilation centered at the origin with scale factor \( k \) is \( (x, y) \to (kx, ky) \). Here, \( k = \frac{1}{4} \).

For \( R' \):

Multiply \( R(-8, -8) \) by \( \frac{1}{4} \):
\( x' = -8 \times \frac{1}{4} = -2 \)
\( y' = -8 \times \frac{1}{4} = -2 \)
So, \( R'(-2, -2) \)

For \( S' \):

Multiply \( S(-8, 8) \) by \( \frac{1}{4} \):
\( x' = -8 \times \frac{1}{4} = -2 \)
\( y' = 8 \times \frac{1}{4} = 2 \)
So, \( S'(-2, 2) \)

For \( T' \):

Multiply \( T(4, 8) \) by \( \frac{1}{4} \):
\( x' = 4 \times \frac{1}{4} = 1 \)
\( y' = 8 \times \frac{1}{4} = 2 \)
So, \( T'(1, 2) \)

For \( U' \):

Multiply \( U(4, -8) \) by \( \frac{1}{4} \):
\( x' = 4 \times \frac{1}{4} = 1 \)
\( y' = -8 \times \frac{1}{4} = -2 \)
So, \( U'(1, -2) \)

Answer:

\( R'(-2, -2) \)
\( S'(-2, 2) \)
\( T'(1, 2) \)
\( U'(1, -2) \)