Question

transform quadrilateral efgh by using the rule $r_{y\text{-axis}} \circ r_{90^\circ}(x, y)$. plot a point for each vertex.

Explanation:

Step1: Identify Original Coordinates

First, find the coordinates of each vertex of quadrilateral EFGH. From the graph:

• \( E(-8, -6) \)
• \( F(-5, -2) \)
• \( G(-2, -3) \)
• \( H(-5, -7) \)

Step2: Apply \( R_{90^\circ} \) Rotation

The rule for a \( 90^\circ \) counterclockwise rotation about the origin \( R_{90^\circ}(x, y) \) is \( (x, y) \to (-y, x) \).

• For \( E(-8, -6) \): \( R_{90^\circ}(-8, -6) = (6, -8) \)
• For \( F(-5, -2) \): \( R_{90^\circ}(-5, -2) = (2, -5) \)
• For \( G(-2, -3) \): \( R_{90^\circ}(-2, -3) = (3, -2) \)
• For \( H(-5, -7) \): \( R_{90^\circ}(-5, -7) = (7, -5) \)

Step3: Apply \( r_{y\text{-axis}} \) Reflection

The rule for reflection over the \( y \)-axis \( r_{y\text{-axis}}(x, y) \) is \( (x, y) \to (-x, y) \).

• For \( E' (6, -8) \) (after \( R_{90^\circ} \)): \( r_{y\text{-axis}}(6, -8) = (-6, -8) \)
• For \( F' (2, -5) \) (after \( R_{90^\circ} \)): \( r_{y\text{-axis}}(2, -5) = (-2, -5) \)
• For \( G' (3, -2) \) (after \( R_{90^\circ} \)): \( r_{y\text{-axis}}(3, -2) = (-3, -2) \)
• For \( H' (7, -5) \) (after \( R_{90^\circ} \)): \( r_{y\text{-axis}}(7, -5) = (-7, -5) \)

Step4: Plot the Transformed Points

Plot the points \( (-6, -8) \), \( (-2, -5) \), \( (-3, -2) \), and \( (-7, -5) \) on the coordinate plane.

Answer:

The transformed vertices are \( E''(-6, -8) \), \( F''(-2, -5) \), \( G''(-3, -2) \), and \( H''(-7, -5) \) (plot these points on the given grid).