Question

reflection across $x = -3$

Explanation:

To solve the reflection of a point across the vertical line \( x = -3 \), we use the formula for reflecting a point \( (x, y) \) across the vertical line \( x = a \): the new \( x \)-coordinate is \( 2a - x \), and the \( y \)-coordinate remains the same. Here, \( a = -3 \), so the formula becomes \( x' = 2(-3) - x = -6 - x \) and \( y' = y \).

Step 1: Identify Coordinates of Points

First, we determine the coordinates of points \( I \), \( P \), \( S \), and \( W \) from the grid:

• \( I \): Let’s assume \( I \) is at \( (-6, -1) \) (by counting grid units: 6 left on \( x \)-axis, 1 down on \( y \)-axis).
• \( P \): \( (-5, -2) \) (5 left, 2 down).
• \( S \): \( (-3, 0) \) (3 left, 0 on \( y \)-axis).
• \( W \): \( (-3, 2) \) (3 left, 2 up).

Step 2: Reflect Each Point Across \( x = -3 \)

Using the reflection formula \( x' = -6 - x \), \( y' = y \):

For Point \( I(-6, -1) \):

\( x' = -6 - (-6) = 0 \), \( y' = -1 \).
Reflected \( I' \): \( (0, -1) \).

For Point \( P(-5, -2) \):

\( x' = -6 - (-5) = -1 \), \( y' = -2 \).
Reflected \( P' \): \( (-1, -2) \).

For Point \( S(-3, 0) \):

\( x' = -6 - (-3) = -3 \), \( y' = 0 \).
Reflected \( S' \): \( (-3, 0) \) (since \( S \) lies on \( x = -3 \), it reflects to itself).

For Point \( W(-3, 2) \):

\( x' = -6 - (-3) = -3 \), \( y' = 2 \).
Reflected \( W' \): \( (-3, 2) \) (since \( W \) lies on \( x = -3 \), it reflects to itself).

Final Reflected Points

The reflected vertices are:

• \( I' \): \( (0, -1) \)
• \( P' \): \( (-1, -2) \)
• \( S' \): \( (-3, 0) \)
• \( W' \): \( (-3, 2) \)

(If the original coordinates were misidentified, adjust based on grid counting. For example, if \( S \) is at \( (-3, 0) \) (on \( x = -3 \)), it remains \( (-3, 0) \) after reflection, confirming the formula works for points on the line of reflection.)

Answer:

The reflected points are \( I'(0, -1) \), \( P'(-1, -2) \), \( S'(-3, 0) \), and \( W'(-3, 2) \). (Plot these on the grid to see the reflected figure.)