QUESTION IMAGE
Question
transform quadrilateral efgh by using the rule ( r_{y\text{-axis}} circ r_{0,90^circ}(x, y) ). plot a point for each vertex.
To solve the transformation \( r_{y\text{-axis}} \circ R_{90^\circ}(x, y) \), we first recall the rules for each transformation:
Step 1: Recall the \( 90^\circ \) rotation rule
A \( 90^\circ \) counterclockwise rotation about the origin (\( R_{90^\circ} \)) transforms a point \( (x, y) \) to \( (-y, x) \).
Step 2: Recall the reflection over the \( y \)-axis rule
A reflection over the \( y \)-axis (\( r_{y\text{-axis}} \)) transforms a point \( (x, y) \) to \( (-x, y) \).
Step 3: Apply the composition \( r_{y\text{-axis}} \circ R_{90^\circ}(x, y) \)
The composition means we first apply \( R_{90^\circ} \), then apply \( r_{y\text{-axis}} \) to the result.
- Let the original point be \( (x, y) \).
- Apply \( R_{90^\circ} \): \( (x, y) \to (-y, x) \).
- Apply \( r_{y\text{-axis}} \) to \( (-y, x) \):
- The \( x \)-coordinate of \( (-y, x) \) is \( -y \), so reflecting over the \( y \)-axis gives \( -(-y) = y \).
- The \( y \)-coordinate remains \( x \).
- Thus, \( r_{y\text{-axis}}(-y, x) = (y, x) \).
Step 4: Identify vertices of \( EFGH \) (from the graph)
Assume the coordinates of \( E, F, G, H \) (from the grid) are:
- \( E(-6, -8) \)
- \( F(-2, -5) \)
- \( G(-2, -2) \)
- \( H(-6, -5) \)
Step 5: Apply the composition to each vertex
For each vertex \( (x, y) \), use \( (x, y) \to (y, x) \):
- Vertex \( E(-6, -8) \):
\( (x, y) = (-6, -8) \to (y, x) = (-8, -6) \).
- Vertex \( F(-2, -5) \):
\( (x, y) = (-2, -5) \to (y, x) = (-5, -2) \).
- Vertex \( G(-2, -2) \):
\( (x, y) = (-2, -2) \to (y, x) = (-2, -2) \).
- Vertex \( H(-6, -5) \):
\( (x, y) = (-6, -5) \to (y, x) = (-5, -6) \).
Step 6: Plot the transformed points
Plot \( (-8, -6) \) (for \( E' \)), \( (-5, -2) \) (for \( F' \)), \( (-2, -2) \) (for \( G' \)), and \( (-5, -6) \) (for \( H' \)) on the grid.
Final Answer
The transformed vertices are \( E'(-8, -6) \), \( F'(-5, -2) \), \( G'(-2, -2) \), and \( H'(-5, -6) \). Plot these points to complete the transformation.
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To solve the transformation \( r_{y\text{-axis}} \circ R_{90^\circ}(x, y) \), we first recall the rules for each transformation:
Step 1: Recall the \( 90^\circ \) rotation rule
A \( 90^\circ \) counterclockwise rotation about the origin (\( R_{90^\circ} \)) transforms a point \( (x, y) \) to \( (-y, x) \).
Step 2: Recall the reflection over the \( y \)-axis rule
A reflection over the \( y \)-axis (\( r_{y\text{-axis}} \)) transforms a point \( (x, y) \) to \( (-x, y) \).
Step 3: Apply the composition \( r_{y\text{-axis}} \circ R_{90^\circ}(x, y) \)
The composition means we first apply \( R_{90^\circ} \), then apply \( r_{y\text{-axis}} \) to the result.
- Let the original point be \( (x, y) \).
- Apply \( R_{90^\circ} \): \( (x, y) \to (-y, x) \).
- Apply \( r_{y\text{-axis}} \) to \( (-y, x) \):
- The \( x \)-coordinate of \( (-y, x) \) is \( -y \), so reflecting over the \( y \)-axis gives \( -(-y) = y \).
- The \( y \)-coordinate remains \( x \).
- Thus, \( r_{y\text{-axis}}(-y, x) = (y, x) \).
Step 4: Identify vertices of \( EFGH \) (from the graph)
Assume the coordinates of \( E, F, G, H \) (from the grid) are:
- \( E(-6, -8) \)
- \( F(-2, -5) \)
- \( G(-2, -2) \)
- \( H(-6, -5) \)
Step 5: Apply the composition to each vertex
For each vertex \( (x, y) \), use \( (x, y) \to (y, x) \):
- Vertex \( E(-6, -8) \):
\( (x, y) = (-6, -8) \to (y, x) = (-8, -6) \).
- Vertex \( F(-2, -5) \):
\( (x, y) = (-2, -5) \to (y, x) = (-5, -2) \).
- Vertex \( G(-2, -2) \):
\( (x, y) = (-2, -2) \to (y, x) = (-2, -2) \).
- Vertex \( H(-6, -5) \):
\( (x, y) = (-6, -5) \to (y, x) = (-5, -6) \).
Step 6: Plot the transformed points
Plot \( (-8, -6) \) (for \( E' \)), \( (-5, -2) \) (for \( F' \)), \( (-2, -2) \) (for \( G' \)), and \( (-5, -6) \) (for \( H' \)) on the grid.
Final Answer
The transformed vertices are \( E'(-8, -6) \), \( F'(-5, -2) \), \( G'(-2, -2) \), and \( H'(-5, -6) \). Plot these points to complete the transformation.