Question

write the coordinates of the vertices after a reflection over the line y = x.
s((□,□))
t((□,□))
u((□,□))
v((□,□))

Explanation:

Step1: Find original coordinates

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

• \( S \): Looking at the grid, \( S \) is at \( (1, -9) \) (wait, no, let's check again. Wait, the vertical line from \( x = 1 \) (since the green line is at \( x = 1 \) and \( x = 9 \)? Wait, no, the grid: the x-axis and y-axis. Let's re-examine. The point \( S \): the green dot at the bottom left of the rectangle. Let's see, the x-coordinate: from the origin (0,0), moving right 1 unit? Wait, no, the vertical line for \( S \) and the other points. Wait, the rectangle has vertices: let's list them properly.

Looking at the graph:

• \( S \): The bottom left vertex. Let's check the coordinates. The x-coordinate: when y = -9 (since it's on the line y = -9), and x = 1? Wait, no, the vertical lines: one at x = 1 (the left vertical line) and x = 9 (the right vertical line)? Wait, the right vertex \( U \) is at (9, 0) (since it's on the x-axis, y=0, x=9). Then \( T \) is at (9, -9) (bottom right, same x as U, y = -9). \( S \) is at (1, -9) (bottom left, same y as T, x=1). \( V \) is at (1, 0) (top left, same x as S, y=0).

So original coordinates:

• \( S(1, -9) \)
• \( T(9, -9) \)
• \( U(9, 0) \)
• \( V(1, 0) \)

Step2: Apply reflection over \( y = x \)

The rule for reflection over the line \( y = x \) is to swap the x and y coordinates. So if a point is \( (a, b) \), its reflection \( (b, a) \).

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

Swap x and y: \( S'(-9, 1) \)

For \( T(9, -9) \):

Swap x and y: \( T'(-9, 9) \)

For \( U(9, 0) \):

Swap x and y: \( U'(0, 9) \)

For \( V(1, 0) \):

Swap x and y: \( V'(0, 1) \)

Wait, wait, maybe I made a mistake in original coordinates. Let's recheck the graph. The x-axis: the rightmost green dot \( U \) is at x=9 (since it's at 9 on the x-axis, y=0). The bottom right \( T \) is at x=9, y=-9 (since it's 9 units down from U). The bottom left \( S \) is at x=1, y=-9 (since it's 8 units left from T? Wait, no, the grid lines: from x=1 to x=9 is 8 units? Wait, no, the grid has each square as 1 unit. Let's count the x-coordinates:

• \( V \): top left, on x=1 (since it's 1 unit right of y-axis), y=0 (on x-axis). So \( V(1, 0) \)
• \( U \): top right, on x=9 (9 units right of y-axis), y=0. So \( U(9, 0) \)
• \( T \): bottom right, same x as U (9), y=-9 (9 units down from U). So \( T(9, -9) \)
• \( S \): bottom left, same x as V (1), y=-9 (9 units down from V). So \( S(1, -9) \)

Yes, that's correct. Now reflection over \( y = x \): swap x and y.

So:

• \( S(1, -9) \) reflects to \( S'(-9, 1) \)
• \( T(9, -9) \) reflects to \( T'(-9, 9) \)
• \( U(9, 0) \) reflects to \( U'(0, 9) \)
• \( V(1, 0) \) reflects to \( V'(0, 1) \)

Wait, but let's verify with the line \( y = x \). The reflection over \( y = x \) swaps x and y, so (a,b) becomes (b,a). So for (1,0), swapping gives (0,1), which is correct. For (9,0), swapping gives (0,9). For (9,-9), swapping gives (-9,9). For (1,-9), swapping gives (-9,1).

Answer:

\( S'(-9, 1) \)
\( T'(-9, 9) \)
\( U'(0, 9) \)
\( V'(0, 1) \)