Question

write the coordinates of the vertices after a reflection over the line x = 1.

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of each vertex from the graph:

• \( B(2, -6) \)
• \( C(9, -6) \) (assuming the grid, since from x=1 to x=9 is 8 units right, but let's confirm: looking at the graph, B is at x=2, y=-6; C is at x=9, y=-6; D is at x=5, y=1; E is at x=3, y=1) Wait, maybe I misread. Let's recheck:

Looking at the graph, the line x=1 is the vertical line. Let's find each point:

• \( B \): x=2, y=-6 (since it's 1 unit right of x=1? Wait, no, the grid: each square is 1 unit. So B is at (2, -6), E at (3, 1), D at (5, 1), C at (9, -6).

Step2: Reflection over x=1

The formula for reflecting a point \( (x, y) \) over the vertical line \( x = a \) is \( (2a - x, y) \). Here, \( a = 1 \), so the reflection is \( (2(1) - x, y) = (2 - x, y) \).

For \( B(2, -6) \):

Step3: Reflect B

Apply the formula: \( x' = 2 - 2 = 0 \), \( y' = -6 \). So \( B'(0, -6) \).

For \( C(9, -6) \):

Step4: Reflect C

\( x' = 2 - 9 = -7 \), \( y' = -6 \). So \( C'(-7, -6) \).

For \( D(5, 1) \):

Step5: Reflect D

\( x' = 2 - 5 = -3 \), \( y' = 1 \). So \( D'(-3, 1) \).

For \( E(3, 1) \):

Step6: Reflect E

\( x' = 2 - 3 = -1 \), \( y' = 1 \). So \( E'(-1, 1) \).

Answer:

\( B'(0, -6) \)
\( C'(-7, -6) \)
\( D'(-3, 1) \)
\( E'(-1, 1) \)