Question

write the coordinates of the vertices after a reflection over the line x = 1. b((box, box)) c((box, box)) d((box, box))

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of points \( B \), \( C \), and \( D \) from the graph.
- For point \( B \): Looking at the grid, \( B \) is at \( (1, 7) \)? Wait, no, wait. Wait, the x - coordinate: let's check the grid. The vertical line is x, horizontal is y. Wait, the original points: Let's see, \( B \) is at \( (1, 7) \)? Wait, no, maybe I misread. Wait, the graph: the y - axis is vertical. Let's check the x - coordinate. Wait, the line \( x = 1 \) is the vertical line. Wait, original coordinates:
- Point \( B \): Let's see, the x - coordinate is 1? Wait, no, looking at the grid, the first vertical line after y - axis (x = 0) is x = 1? Wait, no, the grid lines: each square is 1 unit. So \( B \) is at \( (1, 7) \)? Wait, no, maybe \( B \) is at \( (1, 7) \), \( D \) at \( (1, 6) \), \( C \) at \( (10, 7) \). Wait, let's confirm:
- \( B \): x = 1, y = 7 (since it's on the vertical line x = 1, y = 7)
- \( D \): x = 1, y = 6
- \( C \): x = 10, y = 7

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 formula is \( (2(1)-x, y)=(2 - x, y) \).

For point \( B(1, 7) \):

Substitute \( x = 1 \), \( y = 7 \) into the reflection formula: \( 2 - 1 = 1 \), \( y = 7 \). Wait, that can't be right. Wait, maybe I got the original x - coordinate wrong. Wait, maybe \( B \) is at \( (1, 7) \)? No, wait, maybe the original \( B \) is at \( (1, 7) \)? Wait, no, let's re - examine the graph. Wait, the y - axis is x = 0. Then the first vertical line to the right is x = 1, x = 2, etc. So \( B \) is at \( (1, 7) \), \( D \) at \( (1, 6) \), \( C \) at \( (10, 7) \).

Wait, no, maybe \( B \) is at \( (1, 7) \), \( D \) at \( (1, 6) \), \( C \) at \( (10, 7) \). Now, reflecting over \( x = 1 \):

For a point \( (x, y) \) reflected over \( x = 1 \), the distance from \( x \) to \( x = 1 \) is \( |x - 1| \), so the reflected x - coordinate is \( 1 - (x - 1)=2 - x \), y remains the same.

Reflecting \( B(1, 7) \):

\( x = 1 \), so \( 2 - 1 = 1 \), \( y = 7 \). Wait, that's the same point. That can't be. So I must have misread the original x - coordinate of \( B \). Wait, maybe \( B \) is at \( (1, 7) \)? No, maybe the original \( B \) is at \( (1, 7) \), but that's on the line \( x = 1 \), so reflection is the same. But that seems odd. Wait, maybe the original \( B \) is at \( (0, 7) \)? Wait, no, the graph: the y - axis is x = 0. Let's check again.

Wait, maybe the original coordinates are:
- \( B \): \( (1, 7) \) – no, maybe \( B \) is at \( (1, 7) \), \( D \) at \( (1, 6) \), \( C \) at \( (10, 7) \). Wait, let's check \( C \): x = 10, y = 7. Then reflecting \( C(10, 7) \) over \( x = 1 \): \( 2(1)-10 = 2 - 10=-8 \), \( y = 7 \). So \( C'(-8, 7) \).

For \( B(1, 7) \): \( 2(1)-1 = 1 \), so \( B'(1, 7) \) (since it's on the line of reflection, x = 1, so reflection is itself).

For \( D(1, 6) \): \( 2(1)-1 = 1 \), so \( D'(1, 6) \) (also on the line of reflection). Wait, that can't be right. There must be a mistake in original coordinates.

Wait, maybe the original \( B \) is at \( (0, 7) \), \( D \) at \( (0, 6) \), \( C \) at \( (9, 7) \)? Wait, no, the graph shows \( C \) at x = 10, y = 7. Let's try again.

Wait, the correct way: The line of reflection is \( x = 1 \). For a point \( (x, y) \), reflection over \( x = 1 \) is \( (2\times1 - x, y)=(2 - x, y) \).

Let's assume the original coordinates:
- \( B \): \( (1, 7) \) – then \( B'=(2 - 1, 7)=(1, 7) \)
- \( D \): \( (1, 6) \) – \( D'=(2 - 1, 6)=(1, 6) \)
- \( C \): \( (10, 7) \) – \( C'=(2 - 10, 7)=(-8, 7) \)

Wait, but that seems like \( B \) and \( D \) are on the line of reflection, so their reflections are the same. Maybe that's correct. Let's confirm with the graph. If \( B \) is at \( (1, 7) \), which is on \( x = 1 \), then reflecting over \( x = 1 \) leaves it unchanged. \( C \) is at \( (10, 7) \), so the distance from \( x = 10 \) to \( x = 1 \) is 9 units, so the reflection is 9 units to the left of \( x = 1 \), so \( x = 1 - 9=-8 \), y remains 7. \( D \) is at \( (1, 6) \), on \( x = 1 \), so reflection is \( (1, 6) \).

So:
- \( B'(1, 7) \)
- \( C'(-8, 7) \)
- \( D'(1, 6) \)

Wait, but maybe the original \( B \) is at \( (1, 7) \), \( D \) at \( (1, 6) \), \( C \) at \( (10, 7) \). So the reflections are:

For \( B(1, 7) \): \( x' = 2(1)-1 = 1 \), \( y' = 7 \) → \( B'(1, 7) \)

For \( C(10, 7) \): \( x' = 2(1)-10=-8 \), \( y' = 7 \) → \( C'(-8, 7) \)

For \( D(1, 6) \): \( x' = 2(1)-1 = 1 \), \( y' = 6 \) → \( D'(1, 6) \)

Answer:

\( B'(1, 7) \), \( C'(-8, 7) \), \( D'(1, 6) \)