Question

write the coordinates of the vertices after a reflection over the line $y = -x$.\
$b(\square,\square)$\
$c(\square,\square)$\
$d(\square,\square)$

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of points \( B \), \( C \), and \( D \) from the graph.
- For point \( B \): From the graph, \( B \) is at \( (3, 5) \) (wait, no, looking at the grid: the x - coordinate of \( B \) is 3? Wait, no, the grid lines: let's check again. The y - axis is vertical, x - axis horizontal. Point \( D \) is at \( (0, 4) \), point \( B \): let's see, the x - coordinate is 3? Wait, no, the grid: each square is 1 unit. Let's check the coordinates:
- Point \( D \): \( (0, 4) \) (since it's on the y - axis, x = 0, y = 4)
- Point \( B \): Let's count the x - units from the origin (0,0). Moving right 3 units? Wait, no, the graph: the x - coordinate of \( B \) is 3? Wait, no, looking at the graph, \( B \) is at \( (3, 5) \)? Wait, no, the y - coordinate of \( B \) is 5? Wait, no, the original graph: \( D \) is at (0,4), \( B \) is at (3,5)? Wait, no, let's re - examine. The y - axis: the lines are at y = 0, y = 2, y = 4, y = 6, etc. The x - axis: x = 0, x = 2, x = 4, etc. Wait, point \( B \): x = 3? No, maybe I made a mistake. Wait, the graph: point \( B \) is at (3, 5)? Wait, no, the coordinates: let's look at the grid. The vertical lines (x - axis) and horizontal lines (y - axis). Let's see:

Wait, the original points:

- \( D \): (0, 4) (x = 0, y = 4)
- \( B \): Let's see, from the origin (0,0), moving right 3 units (x = 3) and up 5 units (y = 5)? No, wait, the y - coordinate of \( B \) is 5? Wait, the graph shows \( B \) is at (3, 5)? Wait, no, the y - axis: the line for y = 4, y = 5? Wait, the original \( B \) is at (3, 5)? Wait, no, looking at the graph, \( B \) is at (3, 5)? Wait, maybe I misread. Wait, the problem: the triangle has points \( D \), \( B \), \( C \). \( D \) is at (0,4), \( B \) is at (3,5)? Wait, no, the x - coordinate of \( B \) is 3? Wait, no, the grid: each square is 1 unit. So \( B \) is at (3, 5)? Wait, no, the y - coordinate of \( B \) is 5? Wait, the original \( B \): x = 3, y = 5? \( C \) is at (3, 10)? No, \( C \) is at (3, 10)? Wait, no, the y - coordinate of \( C \) is 10? Wait, the graph: \( C \) is at (3, 10)? Wait, no, the y - axis goes up to 10. So \( C \) is at (3, 10)? Wait, no, the x - coordinate of \( C \) is 3? Wait, no, the x - coordinate of \( C \) is 3? Wait, maybe I made a mistake. Let's start over.

The rule for reflection over \( y=-x \) is: if a point has coordinates \( (x,y) \), its reflection over \( y = -x \) is \( (-y,-x) \).

So first, find the original coordinates:

- \( D \): (0, 4)
- \( B \): Let's look at the graph again. The x - coordinate of \( B \): from the origin, moving right 3 units (x = 3), y - coordinate: moving up 5 units (y = 5)? Wait, no, the y - coordinate of \( B \) is 5? Wait, the graph shows \( B \) is at (3, 5)? Wait, no, the y - axis: the line for y = 4, y = 5? Wait, the original \( B \) is at (3, 5)? Wait, maybe. Let's confirm:

- \( D \): (0, 4) (x = 0, y = 4)
- \( B \): (3, 5) (x = 3, y = 5)
- \( C \): (3, 10) (x = 3, y = 10)

Now, the rule for reflection over \( y=-x \) is \( (x,y)\to(-y,-x) \)

Step2: Reflect \( D \)

For point \( D=(0,4) \):

Using the reflection rule \( (x,y)\to(-y,-x) \), substitute \( x = 0 \), \( y = 4 \)

\( D'=(-4,0) \) (since \( -y=-4 \), \( -x = 0 \))

Step3: Reflect \( B \)

For point \( B=(3,5) \):

Using the rule \( (x,y)\to(-y,-x) \), substitute \( x = 3 \), \( y = 5 \)

\( B'=(-5,-3) \) (since \( -y=-5 \), \( -x=-3 \))

Step4: Reflect \( C \)

For point \( C=(3,10) \):

Using the rule \( (x,y)\to(-y,-x) \), substitute \( x = 3 \), \( y = 10 \)

\( C'=(-10,-3) \) (since \( -y=-10 \), \( -x=-3 \))

Wait, maybe I made a mistake in the original coordinates. Let's re - check the original coordinates.

Wait, the graph:

- \( D \): (0, 4) (correct, on y - axis, x = 0, y = 4)
- \( B \): Let's count the x - units. From the origin (0,0), moving right 3 units (x = 3) and up 5 units (y = 5)? Wait, no, the y - coordinate of \( B \) is 5? Wait, the graph shows \( B \) is at (3, 5)? Wait, no, the y - axis: the line for y = 4, y = 5? Wait, the original \( B \) is at (3, 5)? Wait, maybe the x - coordinate of \( B \) is 3 and y - coordinate is 5. Then reflection over \( y=-x \) is \( (-y,-x) \), so \( B'=(-5,-3) \)

- \( C \): x = 3, y = 10. So reflection is \( (-10,-3) \)

- \( D \): x = 0, y = 4. Reflection is \( (-4,0) \)

Wait, maybe the original coordinates are:

- \( D \): (0, 4)
- \( B \): (3, 5)
- \( C \): (3, 10)

So applying the reflection rule \( (x,y)\to(-y,-x) \):

- For \( B(3,5) \): \( B'(-5,-3) \)
- For \( C(3,10) \): \( C'(-10,-3) \)
- For \( D(0,4) \): \( D'(-4,0) \)

Wait, maybe I made a mistake in the x - coordinate of \( B \) and \( C \). Let's re - check the graph. The x - coordinate of \( B \) and \( C \) is 3? Because they are vertically aligned (same x - coordinate). So \( B \) is (3, 5) and \( C \) is (3, 10), \( D \) is (0, 4).

Yes, that makes sense. So the reflection over \( y=-x \) is given by the transformation \( (x,y)\to(-y,-x) \).

So:

- \( B(3,5)\to B'(-5,-3) \)
- \( C(3,10)\to C'(-10,-3) \)
- \( D(0,4)\to D'(-4,0) \)

Answer:

\( B'(-5, -3) \)
\( C'(-10, -3) \)
\( D'(-4, 0) \)