Question

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

Explanation:

Step1: Find original coordinates

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

• \( C \): Looking at the graph, \( C \) is at \( (3, -1) \)? Wait, no, wait. Wait, the grid: Let's check again. Wait, the x-axis: \( C \) is at x=3? Wait, no, the point C is at (3, -1)? Wait, no, looking at the graph, the purple line is y=-1? Wait, no, the original points: Let's see, D is at (3,7)? Wait, no, the y-axis: each grid is 1 unit. Let's re-express:

Wait, the original coordinates:

• \( C \): From the graph, \( C \) is at \( (3, -1) \)? Wait, no, the point C is on the line y=-1? Wait, no, the purple line is y=-1? Wait, no, the x-axis is the horizontal line, and the y-axis vertical. Let's check the coordinates:

Looking at the graph:

• \( C \): x=3, y=-1? Wait, no, the point C is at (3, -1)? Wait, no, the blue line from D to C: D is at (3,7), C is at (3, -1)? Wait, no, the y-coordinate of C: looking at the grid, the horizontal lines: the line y=0 is the x-axis, then y=-1 is one unit below. So \( C \) is at \( (3, -1) \)? Wait, no, the original C: let's check the graph again. Wait, the point C is at (3, -1)? Wait, maybe I made a mistake. Wait, the original points:

Wait, D is at (3,7), E is at (8,8), F is at (8,2), C is at (3, -1)? Wait, no, the line from C to D is vertical, so same x-coordinate. So D is (3,7), C is (3, -1). Then F is (8,2), E is (8,8).

Step2: Reflection over \( y = -1 \)

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

For \( C(3, -1) \):

Apply the formula: \( x = 3 \), \( y = -1 \). So \( y' = -2 - (-1) = -2 + 1 = -1 \). So \( C'(3, -1) \). Wait, that's because C is on the line \( y = -1 \), so reflecting over the line leaves it unchanged.

For \( D(3, 7) \):

\( x = 3 \), \( y = 7 \). \( y' = -2 - 7 = -9 \). So \( D'(3, -9) \).

For \( E(8, 8) \):

\( x = 8 \), \( y = 8 \). \( y' = -2 - 8 = -10 \). So \( E'(8, -10) \).

For \( F(8, 2) \):

\( x = 8 \), \( y = 2 \). \( y' = -2 - 2 = -4 \). So \( F'(8, -4) \).

Wait, let's verify the reflection formula. The distance from the point to the line \( y = k \) is \( |y - k| \). So the reflected point should be on the other side of the line, at the same distance. So for a point \( (x, y) \), the distance to \( y = -1 \) is \( |y - (-1)| = |y + 1| \). So the reflected y-coordinate is \( -1 - |y + 1| \) if \( y > -1 \), or \( -1 + |y + 1| \) if \( y < -1 \). Let's check with D(3,7): distance is \( |7 - (-1)| = 8 \). So reflected y-coordinate is \( -1 - 8 = -9 \), which matches. For F(8,2): distance is \( |2 - (-1)| = 3 \), so reflected y-coordinate is \( -1 - 3 = -4 \), which matches. For E(8,8): distance is \( |8 - (-1)| = 9 \), reflected y-coordinate \( -1 - 9 = -10 \), matches. For C(3, -1): distance is \( |-1 - (-1)| = 0 \), so reflected y-coordinate is \( -1 - 0 = -1 \), so C' is (3, -1).

Answer:

\( C'(3, -1) \)
\( D'(3, -9) \)
\( E'(8, -10) \)
\( F'(8, -4) \)