Question

write the coordinates of the vertices after a reflection over the y-axis.
k(□, □)
l(□, □)
m(□, □)

Explanation:

Step1: Find original coordinates

From the graph, we identify the original coordinates:

• \( K \): Looking at the grid, \( K \) is at \( (9, -1) \) (wait, no, let's check again. Wait, the x-axis: K is at x=9? Wait, no, the grid lines. Wait, K is at (9, -1)? Wait, no, looking at the graph, K is at (9, -1)? Wait, no, the x-coordinate: from the origin, moving right 9? Wait, no, the grid has x from -10 to 10, y from -10 to 10. Let's re-examine:

Wait, K is at (9, -1)? Wait, no, the blue dot for K: x=9, y=-1? Wait, no, the y-axis: below the x-axis is negative. So K is (9, -1)? Wait, L is at (9, 5) (since y=5, x=9). M is at (6, -2) (x=6, y=-2). Let's confirm:

• \( K \): x=9, y=-1 (since it's on the x=9 line, y=-1)
• \( L \): x=9, y=5 (on x=9, y=5)
• \( M \): x=6, y=-2 (on x=6, y=-2)

Step2: Apply reflection over y-axis

The rule for reflection over the y-axis is \( (x, y)
ightarrow (-x, y) \).

For \( K(9, -1) \):

Apply the rule: \( x = 9
ightarrow -9 \), \( y = -1 \) remains. So \( K'(-9, -1) \).

For \( L(9, 5) \):

Apply the rule: \( x = 9
ightarrow -9 \), \( y = 5 \) remains. So \( L'(-9, 5) \).

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

Apply the rule: \( x = 6
ightarrow -6 \), \( y = -2 \) remains. So \( M'(-6, -2) \).

Answer:

\( K'(-9, -1) \), \( L'(-9, 5) \), \( M'(-6, -2) \)

Wait, wait, maybe I made a mistake in original coordinates. Let's recheck:

Looking at the graph:

• K: x=9, y=-1? Wait, no, the y-coordinate: the point K is on the x-axis? No, below the x-axis. Wait, the grid lines: each square is 1 unit. So K is at (9, -1)? Wait, L is at (9, 5) (since from x=9, y=5). M is at (6, -2) (x=6, y=-2). Yes. So reflection over y-axis: (x,y) → (-x,y). So:

K: (9, -1) → (-9, -1)

L: (9, 5) → (-9, 5)

M: (6, -2) → (-6, -2)

Yes, that's correct.