Question

graph the image of parallelogram jklm after a reflection over the line $y = -x$.

Explanation:

Step1: Identify coordinates of J, K, L, M

First, find the coordinates of the vertices of parallelogram \( JKLM \). From the graph:

• \( J(-9, -10) \)
• \( K(-5, -10) \)
• \( L(2, -8) \)
• \( M(-2, -8) \)

Step2: Apply reflection over \( y = -x \)

The rule for reflecting a point \( (x, y) \) over the line \( y = -x \) is \( (x, y) \to (-y, -x) \).

• For \( J(-9, -10) \):

Apply the rule: \( (-(-10), -(-9)) = (10, 9) \)

• For \( K(-5, -10) \):

Apply the rule: \( (-(-10), -(-5)) = (10, 5) \)

• For \( L(2, -8) \):

Apply the rule: \( (-(-8), -2) = (8, -2) \)

• For \( M(-2, -8) \):

Apply the rule: \( (-(-8), -(-2)) = (8, 2) \)

Step3: Plot the reflected points

Plot the points \( J'(10, 9) \), \( K'(10, 5) \), \( L'(8, -2) \), \( M'(8, 2) \) on the coordinate plane and connect them to form the reflected parallelogram.

Answer:

The reflected parallelogram has vertices at \( J'(10, 9) \), \( K'(10, 5) \), \( L'(8, -2) \), \( M'(8, 2) \) (plotted on the graph as per the reflection rule).