Question

1. parallelogram rstu in the line y = x
2. square klmn in the line y = -2

Explanation:

Step1: Identify Coordinates of Parallelogram RSTU

First, find the coordinates of vertices \( R, S, T, U \) from the grid. Let's assume each grid square has side length 1.

• \( R(-2, 2) \) (2 units left of origin on x, 2 up on y)
• \( S(2, 3) \) (2 right, 3 up)
• \( T(2, -2) \) (2 right, 2 down)
• \( U(-2, -3) \) (2 left, 3 down)

Step2: Reflect Over Line \( y = x \)

The reflection of a point \( (a, b) \) over \( y = x \) is \( (b, a) \). Apply this to each vertex:

• \( R(-2, 2) \to R'(2, -2) \)
• \( S(2, 3) \to S'(3, 2) \)
• \( T(2, -2) \to T'(-2, 2) \)
• \( U(-2, -3) \to U'(-3, -2) \)

Step3: Plot Reflected Points

Plot \( R', S', T', U' \) on the grid and connect them to form the reflected parallelogram.

For Square KLMN (Problem 4, Reflection Over \( y = -2 \)):

Step1: Identify Coordinates of Square KLMN

Find coordinates of \( K, L, M, N \):

• \( K(-1, 0) \) (1 left, 0 on y)
• \( L(-2, 2) \) (2 left, 2 up)
• \( M(1, 3) \) (1 right, 3 up)
• \( N(2, 1) \) (2 right, 1 up)

Step2: Reflect Over Horizontal Line \( y = -2 \)

The reflection of a point \( (a, b) \) over a horizontal line \( y = k \) is \( (a, 2k - b) \). Here, \( k = -2 \), so the formula becomes \( (a, -4 - b) \).

• \( K(-1, 0) \to K'(-1, -4 - 0) = (-1, -4) \)
• \( L(-2, 2) \to L'(-2, -4 - 2) = (-2, -6) \)
• \( M(1, 3) \to M'(1, -4 - 3) = (1, -7) \)
• \( N(2, 1) \to N'(2, -4 - 1) = (2, -5) \)

Step3: Plot Reflected Points

Plot \( K', L', M', N' \) and connect to form the reflected square.

Answer:

For problem 3 (parallelogram over \( y = x \)): Reflected vertices are \( R'(2, -2) \), \( S'(3, 2) \), \( T'(-2, 2) \), \( U'(-3, -2) \).
For problem 4 (square over \( y = -2 \)): Reflected vertices are \( K'(-1, -4) \), \( L'(-2, -6) \), \( M'(1, -7) \), \( N'(2, -5) \).

(Note: The final answer includes the reflected coordinates for both figures. To visualize, plot these points on the grid and connect them.)