Question

write the coordinates of the vertices after a reflection over the line y = 1.
j(□,□)
k(□,□)
l(□,□)
m(□,□)

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of each vertex:

• \( J(-8, -7) \) (since it's at x = -8, y = -7)
• \( K(-7, 0) \)? Wait, no, looking at the graph: M is at (-9, 0), L at (-8, 1), K at (-7, 0), J at (-8, -7). Wait, let's correct:

Wait, the grid: M is at (-9, 0), L at (-8, 1), K at (-7, 0), J at (-8, -7). Wait, no, the x-axis: M is at x=-9, y=0; L at x=-8, y=1; K at x=-7, y=0; J at x=-8, y=-7.

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 new y-coordinate is \( 2(1) - y = 2 - y \), x-coordinate remains the same.

For \( J(-8, -7) \):

New y-coordinate: \( 2 - (-7) = 2 + 7 = 9 \). So \( J'(-8, 9) \)? Wait, no, wait original J: looking at the graph, J is at (-8, -7)? Wait, the y-axis: from 0 down, J is at y=-7? Wait, the grid lines: each square is 1 unit. So J is at x=-8, y=-7? Wait, no, the vertical line: x=-8, and y: from 0 down to -7? Wait, the graph shows J at (-8, -7)? Wait, no, let's check again. The original J: the point J is at x=-8, and y: the distance from y=1 to J: let's see, reflection over y=1.

Wait, maybe I misread J's y-coordinate. Let's look at the graph: J is at (-8, -7)? Wait, the y-axis: 0 is the x-axis, then down to -2, -4, -6, -8. So J is at (-8, -7)? Wait, no, the blue line: J is at (-8, -7)? Wait, no, the vertical line x=-8, and J is at the bottom, y=-7? Wait, no, the grid: each horizontal line is 1 unit. So from y=1 (L's y), down to J: the distance from y=1 to J is 1 - (-7) = 8? Wait, no, reflection over y=1: the distance from the point to y=1 is \( |y - 1| \), so the reflected point is on the other side of y=1, same distance.

So for a point \( (x, y) \), reflection over \( y = 1 \) is \( (x, 1 + (1 - y)) = (x, 2 - y) \).

Let's re-identify the original points:

• M: (-9, 0)
• L: (-8, 1)
• K: (-7, 0)
• J: (-8, -7)

Now apply the reflection:

For J(-8, -7):

New y: \( 2 - (-7) = 9 \). So \( J'(-8, 9) \)? Wait, that can't be, because the graph's y-axis only goes up to 10. Wait, maybe I made a mistake in J's original y-coordinate. Wait, looking at the graph: J is at (-8, -7)? Wait, no, the vertical line x=-8, and J is at y=-7? Wait, the distance from y=1 to J: y=1 to y=-7 is 8 units down, so reflection would be 8 units up from y=1, so y=1 + 8 = 9. So J'(-8, 9).

For K(-7, 0):

New y: \( 2 - 0 = 2 \). So \( K'(-7, 2) \)? Wait, no, K is at (-7, 0)? Wait, M is at (-9, 0), L at (-8, 1), K at (-7, 0), so K is (-7, 0). Then reflection over y=1: y=2 - 0 = 2. So K'(-7, 2).

For L(-8, 1):

New y: \( 2 - 1 = 1 \). So \( L'(-8, 1) \) (since it's on the line y=1, reflection is itself).

For M(-9, 0):

New y: \( 2 - 0 = 2 \). So \( M'(-9, 2) \).

Wait, but maybe I misread the original coordinates. Let's check again:

Looking at the graph:

• M: x=-9, y=0 (on x-axis)
• L: x=-8, y=1 (above x-axis, at y=1)
• K: x=-7, y=0 (on x-axis)
• J: x=-8, y=-7 (below x-axis, at y=-7)

Yes, that's correct. So applying the reflection formula \( (x, 2*1 - y) \):

• J(-8, -7): \( x=-8 \), \( y' = 2 - (-7) = 9 \) → \( J'(-8, 9) \)
• K(-7, 0): \( y' = 2 - 0 = 2 \) → \( K'(-7, 2) \)
• L(-8, 1): \( y' = 2 - 1 = 1 \) → \( L'(-8, 1) \)
• M(-9, 0): \( y' = 2 - 0 = 2 \) → \( M'(-9, 2) \)

Wait, but the problem's J is written as J', K', L', M'. Let's confirm:

Original J: (-8, -7) → reflection over y=1: ( -8, 2 - (-7) ) = (-8, 9)

Original K: (-7, 0) → ( -7, 2 - 0 ) = (-7, 2)

Original L: (-8, 1) → ( -8, 2 - 1 ) = (-8, 1)

Original M: (-9, 0) → ( -9, 2 - 0 ) = (-9, 2)

Yes, that makes…

Answer:

\( J'(-8, 9) \)

\( K'(-7, 2) \)

\( L'(-8, 1) \)

\( M'(-9, 2) \)

Wait, but the problem's J is at (-8, -7)? Wait, maybe I made a mistake in J's y-coordinate. Let's check the graph again. The vertical line x=-8, and J is at the bottom, below the x-axis. The distance from y=1 to J: y=1 to y=0 is 1 unit, then from y=0 down to J: how many units? Let's count the grid lines: from y=1 (L) down to y=0 (M, K) is 1 unit, then down to y=-1, -2, -3, -4, -5, -6, -7. So J is at y=-7. So reflection over y=1: 2*1 - (-7) = 9. So J'(-8, 9).

Yes, that's correct.