Question

write the coordinates of the vertices after a rotation 270° counterclockwise around the origin.
p((□,□))
q((□,□))
r((□,□))

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of points \( P \), \( Q \), and \( R \) from the graph.

• For point \( P \): Looking at the grid, the \( x \)-coordinate is \( -8 \) and the \( y \)-coordinate is \( 3 \), so \( P(-8, 3) \).
• For point \( Q \): The \( x \)-coordinate is \( -8 \) and the \( y \)-coordinate is \( 10 \), so \( Q(-8, 10) \).
• For point \( R \): The \( x \)-coordinate is \( -4 \) and the \( y \)-coordinate is \( 2 \), so \( R(-4, 2) \).

Step2: Apply 270° counterclockwise rotation rule

The rule for rotating a point \( (x, y) \) 270° counterclockwise around the origin is \( (x, y) \to (y, -x) \). This is because a 270° counterclockwise rotation is equivalent to a 90° clockwise rotation, and the transformation for 90° clockwise (or 270° counterclockwise) is \( (x, y) \mapsto (y, -x) \).

For \( P(-8, 3) \):

Using the rule \( (x, y) \to (y, -x) \), substitute \( x = -8 \) and \( y = 3 \).
So, \( x' = 3 \) and \( y' = -(-8) = 8 \). Thus, \( P' (3, 8) \).

For \( Q(-8, 10) \):

Substitute \( x = -8 \) and \( y = 10 \) into the rule \( (x, y) \to (y, -x) \).
\( x' = 10 \) and \( y' = -(-8) = 8 \). Thus, \( Q' (10, 8) \).

For \( R(-4, 2) \):

Substitute \( x = -4 \) and \( y = 2 \) into the rule \( (x, y) \to (y, -x) \).
\( x' = 2 \) and \( y' = -(-4) = 4 \). Thus, \( R' (2, 4) \).

Answer:

\( P'(3, 8) \)
\( Q'(10, 8) \)
\( R'(2, 4) \)