Question

write the coordinates of the vertices after a reflection over the line ( y = -x ).

( j(square, square) )
( k(square, square) )
( l(square, square) )
( m(square, square) )

Explanation:

Step1: Identify original coordinates

First, find the original coordinates of each vertex:
- \( J(-10, -10) \)
- \( K(-10, -7) \) (Wait, looking at the graph, K is at (-10, -7)? Wait no, the grid: Let's recheck. The purple figure: J is at (-10, -10), K is at (-10, -7)? Wait no, the y-axis for K: looking at the grid, the vertical line for K is x=-10, and y-coordinate: from the graph, K is at (-10, -7)? Wait no, the figure is a rectangle. Let's see: J is at (-10, -10), K is at (-10, -7) (since it's vertical up from J), L is at (0, -7)? Wait no, the x-coordinate for L: looking at the grid, L is at (0, -7)? Wait no, the horizontal line from K to L: K is at (-10, -7), L is at (0, -7)? Wait no, the y-coordinate for L: the grid has L at (0, -7)? Wait no, the original coordinates: Let's look again. The purple rectangle: J is at (-10, -10), K is at (-10, -7) (up 3 units), L is at (0, -7) (right 10 units from K), M is at (0, -10) (down 3 units from L). Wait, no, the graph: J is at (-10, -10), K is at (-10, -7), L is at (0, -7), M is at (0, -10). Let's confirm:

- J: x=-10, y=-10 (bottom left)
- K: x=-10, y=-7 (top left)
- L: x=0, y=-7 (top right)
- M: x=0, y=-10 (bottom right)

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

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

For \( J(-10, -10) \):

Apply the rule: \( x = -10 \), \( y = -10 \)
New coordinates: \( (-(-10), -(-10)) = (10, 10) \)? Wait no, wait the rule is \( (x, y) ightarrow (-y, -x) \). Wait, let's recall: reflection over \( y = -x \) swaps and negates the coordinates. So \( (x, y) \) becomes \( (-y, -x) \).

So for \( J(-10, -10) \):
\( x = -10 \), \( y = -10 \)
New \( x' = -y = -(-10) = 10 \)
New \( y' = -x = -(-10) = 10 \)
So \( J' (10, 10) \)? Wait that can't be. Wait maybe I messed up the original coordinates. Wait let's re-express the original coordinates correctly.

Wait looking at the graph: The purple rectangle: J is at (-10, -10), K is at (-10, -7) (since from J, moving up 3 units: y from -10 to -7), L is at (0, -7) (moving right 10 units from K: x from -10 to 0), M is at (0, -10) (moving down 3 units from L: y from -7 to -10). So original coordinates:

- \( J(-10, -10) \)
- \( K(-10, -7) \)
- \( L(0, -7) \)
- \( M(0, -10) \)

Now apply reflection over \( y = -x \): \( (x, y) ightarrow (-y, -x) \)

For \( J(-10, -10) \):

\( x = -10 \), \( y = -10 \)
\( x' = -y = -(-10) = 10 \)
\( y' = -x = -(-10) = 10 \)
So \( J'(10, 10) \)

For \( K(-10, -7) \):

\( x = -10 \), \( y = -7 \)
\( x' = -y = -(-7) = 7 \)
\( y' = -x = -(-10) = 10 \)
So \( K'(7, 10) \)

For \( L(0, -7) \):

\( x = 0 \), \( y = -7 \)
\( x' = -y = -(-7) = 7 \)
\( y' = -x = -0 = 0 \)
So \( L'(7, 0) \)

For \( M(0, -10) \):

\( x = 0 \), \( y = -10 \)
\( x' = -y = -(-10) = 10 \)
\( y' = -x = -0 = 0 \)
So \( M'(10, 0) \)

Wait, but let's verify the reflection rule. The reflection over \( y = -x \) is indeed \( (x, y) \mapsto (-y, -x) \). Let's take a point (a, b), the line \( y = -x \) has slope -1. The midpoint between (a, b) and its reflection (a', b') should lie on \( y = -x \), and the line connecting them should be perpendicular to \( y = -x \) (slope 1).

For example, take J(-10, -10). Let reflection be (x', y'). Midpoint: \( (\frac{-10 + x'}{2}, \frac{-10 + y'}{2}) \) should lie on \( y = -x \), so \( \frac{-10 + y'}{2} = -\frac{-10 + x'}{2} \) → \( -10 + y' = 10 - x' \) → \( x' + y' = 20 \). The slope of JJ' should be 1 (perpendicular to -1), so \( \frac{y' - (-10)}{x' - (-10)} = 1 \) → \( y' + 10 = x' + 10 \) → \( y' = x' \). So from \( x' + y' = 20 \) and \( y' = x' \), we get \( 2x' = 20 \) → \( x' = 10 \), \( y' = 10 \). Which matches our earlier result. Good.

For K(-10, -7): midpoint \( (\frac{-10 + x'}{2}, \frac{-7 + y'}{2}) \) on \( y = -x \): \( \frac{-7 + y'}{2} = -\frac{-10 + x'}{2} \) → \( -7 + y' = 10 - x' \) → \( x' + y' = 17 \). Slope of KK' is 1: \( \frac{y' - (-7)}{x' - (-10)} = 1 \) → \( y' + 7 = x' + 10 \) → \( y' = x' + 3 \). Substitute into \( x' + y' = 17 \): \( x' + x' + 3 = 17 \) → \( 2x' = 14 \) → \( x' = 7 \), \( y' = 10 \). Which matches our earlier result. Good.

For L(0, -7): midpoint \( (\frac{0 + x'}{2}, \frac{-7 + y'}{2}) \) on \( y = -x \): \( \frac{-7 + y'}{2} = -\frac{0 + x'}{2} \) → \( -7 + y' = -x' \) → \( x' + y' = 7 \). Slope of LL' is 1: \( \frac{y' - (-7)}{x' - 0} = 1 \) → \( y' + 7 = x' \) → \( y' = x' - 7 \). Substitute into \( x' + y' = 7 \): \( x' + x' - 7 = 7 \) → \( 2x' = 14 \) → \( x' = 7 \), \( y' = 0 \). Correct.

For M(0, -10): midpoint \( (\frac{0 + x'}{2}, \frac{-10 + y'}{2}) \) on \( y = -x \): \( \frac{-10 + y'}{2} = -\frac{0 + x'}{2} \) → \( -10 + y' = -x' \) → \( x' + y' = 10 \). Slope of MM' is 1: \( \frac{y' - (-10)}{x' - 0} = 1 \) → \( y' + 10 = x' \) → \( y' = x' - 10 \). Substitute into \( x' + y' = 10 \): \( x' + x' - 10 = 10 \) → \( 2x' = 20 \) → \( x' = 10 \), \( y' = 0 \). Correct.

So the reflected coordinates are:

- \( J'(10, 10) \)
- \( K'(7, 10) \)
- \( L'(7, 0) \)
- \( M'(10, 0) \)

Wait, but the problem shows J, K, L, M? Wait the original figure: J, K, L, M. Let's confirm the original coordinates again. Looking at the graph:

- J: x=-10, y=-10 (bottom left)
- K: x=-10, y=-7 (top left)
- L: x=0, y=-7 (top right)
- M: x=0, y=-10 (bottom right)

Yes, that's correct. So applying the reflection rule \( (x, y) ightarrow (-y, -x) \):

- J(-10, -10) → (-(-10), -(-10)) = (10, 10)
- K(-10, -7) → (-(-7), -(-10)) = (7, 10)
- L(0, -7) → (-(-7), -0) = (7, 0)
- M(0, -10) → (-(-10), -0) = (10, 0)

Answer:

• \( J'(10, 10) \)
• \( K'(7, 10) \)
• \( L'(7, 0) \)
• \( M'(10, 0) \)