Question

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

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of \( T \), \( U \), and \( V \) from the graph.

• \( U \) is at \( (3, 0) \) (wait, looking at the graph, \( U \) is at \( (3, 0) \)? Wait, no, the grid: let's check again. Wait, the x-axis: \( U \) is at x=3, y=0? Wait, no, the blue dot for \( U \) is at (3, 0)? Wait, no, looking at the graph, \( U \) is at (3, 0)? Wait, \( T \): let's see, \( T \) is at (3, -6)? Wait, no, the coordinates: let's check the grid. Each square is 1 unit. So:
• \( U \): x=3, y=0 (since it's on the x-axis, 3 units right of origin, y=0)
• \( T \): x=3, y=-6 (3 units right, 6 units down)
• \( V \): x=5, y=-8 (5 units right, 8 units down)

Wait, maybe I misread. Let's re-express:

Looking at the graph:

• \( U \): (3, 0) (x=3, y=0)
• \( T \): (3, -6) (x=3, y=-6)
• \( V \): (5, -8) (x=5, y=-8)

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

The rule for reflection over the line \( y = x \) is \( (x, y)
ightarrow (y, x) \). So we swap the x and y coordinates.

For \( T(3, -6) \):

Swap x and y: \( T'(-6, 3) \)

For \( U(3, 0) \):

Swap x and y: \( U'(0, 3) \)

For \( V(5, -8) \):

Swap x and y: \( V'(-8, 5) \)

Wait, wait, maybe I made a mistake in original coordinates. Let's check again. Let's look at the graph:

• \( U \): the blue dot is at (3, 0)? Wait, the x-axis is horizontal, y-axis vertical. So \( U \) is at (3, 0) (x=3, y=0). \( T \): below \( U \), same x? Wait, no, \( T \) is at (3, -6)? Wait, the vertical line from \( U \) down to \( T \): yes, x=3, y=-6. \( V \): from \( T \), moving right and down: x=5, y=-8? Wait, no, maybe \( T \) is (3, -6), \( V \) is (5, -8)? Wait, let's confirm the grid. Each square is 1 unit. So:

Original coordinates:

• \( U \): (3, 0)
• \( T \): (3, -6)
• \( V \): (5, -8)

Reflection over \( y = x \): \( (x, y) \mapsto (y, x) \)

So:

• \( T(3, -6) \) becomes \( T'(-6, 3) \)
• \( U(3, 0) \) becomes \( U'(0, 3) \)
• \( V(5, -8) \) becomes \( V'(-8, 5) \)

Wait, but maybe I misread the original coordinates. Let's check again. Maybe \( U \) is (3, 0), \( T \) is (3, -6), \( V \) is (5, -8). Let's confirm the reflection rule. Yes, reflection over \( y = x \) swaps x and y. So that's the rule.

Answer:

\( T'(-6, 3) \), \( U'(0, 3) \), \( V'(-8, 5) \)

Wait, but maybe the original coordinates are different. Let's re-express:

Wait, looking at the graph again:

• \( U \): x=3, y=0 (correct, on x-axis)
• \( T \): x=3, y=-6 (correct, 6 units down from U)
• \( V \): x=5, y=-8 (correct, 2 units right and 2 units down from T)

So reflection over \( y = x \):

\( T(3, -6)
ightarrow T'(-6, 3) \)

\( U(3, 0)
ightarrow U'(0, 3) \)

\( V(5, -8)
ightarrow V'(-8, 5) \)

Yes, that's correct.