Question

1. square klmn in the line y = -2

Explanation:

Step1: Identify Coordinates of Vertices

First, find the coordinates of square \( KLMN \):

• \( K(-1, 0) \)
• \( L(-2, 2) \)
• \( M(1, 3) \)
• \( N(2, 1) \)

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

To reflect a point \( (x, y) \) over the horizontal line \( y = k \), the formula is \( (x, 2k - y) \). Here, \( k = -2 \), so the reflection formula becomes \( (x, 2(-2) - y) = (x, -4 - y) \).

Reflect \( K(-1, 0) \):

\( x = -1 \), \( y = 0 \)
New \( y \)-coordinate: \( -4 - 0 = -4 \)
Reflected \( K'(-1, -4) \)

Reflect \( L(-2, 2) \):

\( x = -2 \), \( y = 2 \)
New \( y \)-coordinate: \( -4 - 2 = -6 \)
Reflected \( L'(-2, -6) \)

Reflect \( M(1, 3) \):

\( x = 1 \), \( y = 3 \)
New \( y \)-coordinate: \( -4 - 3 = -7 \)
Reflected \( M'(1, -7) \)

Reflect \( N(2, 1) \):

\( x = 2 \), \( y = 1 \)
New \( y \)-coordinate: \( -4 - 1 = -5 \)
Reflected \( N'(2, -5) \)

Step3: Plot Reflected Points

Plot \( K'(-1, -4) \), \( L'(-2, -6) \), \( M'(1, -7) \), \( N'(2, -5) \) on the grid to get the reflected square.

Answer:

The reflected square \( K'L'M'N' \) has vertices at \( K'(-1, -4) \), \( L'(-2, -6) \), \( M'(1, -7) \), and \( N'(2, -5) \) (plotted by applying the reflection formula \( (x, -4 - y) \) to each vertex of \( KLMN \)).