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 points B, C, D, E from the graph:

• Point B: From the graph, B is at \((-8, -10)\)
• Point C: C is at \((1, -10)\) (Wait, no, looking at the grid, D is at (1, -1)? Wait, no, let's re - examine. The purple figure: E is at (-8, -1), B is at (-8, -10), C is at (1, -10), D is at (1, -1). Wait, the y - axis: the bottom is - 10, so B: (-8, -10), C: (1, -10), D: (1, -1), E: (-8, -1).

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

The rule for reflecting a point \((x,y)\) over the line \(y = -x\) is \((x,y)\to(-y,-x)\).

For point B \((-8,-10)\):

Using the rule \((x,y)\to(-y,-x)\), substitute \(x=-8\) and \(y = - 10\).
We get \(-y=-(-10)=10\) and \(-x=-(-8)=8\). So \(B'=(10,8)\).

For point C \((1,-10)\):

Substitute \(x = 1\) and \(y=-10\) into the rule \((x,y)\to(-y,-x)\).
\(-y=-(-10)=10\) and \(-x=-1\). So \(C'=(10,-1)\).

For point D \((1,-1)\):

Substitute \(x = 1\) and \(y=-1\) into the rule \((x,y)\to(-y,-x)\).
\(-y=-(-1)=1\) and \(-x=-1\). So \(D'=(1,-1)\)? Wait, no, wait: \((x,y)=(1, - 1)\), then \(-y = 1\), \(-x=-1\), so \((1,-1)\) reflected over \(y=-x\) is \((1,-1)\)? Wait, no, let's recalculate. The formula for reflection over \(y=-x\) is \((x,y)\to(-y,-x)\). So for \((1,-1)\): \(-y=-(-1)=1\), \(-x=-1\), so \((1, - 1)\) becomes \((1,-1)\)? Wait, that can't be right. Wait, maybe I made a mistake in original coordinates. Let's re - check the graph. The purple figure: E is at (-8, -1) (since it's on the horizontal line \(y=-1\) and \(x = - 8\)), B is at (-8, -10) (on \(y=-10\), \(x=-8\)), C is at (1, -10) (on \(y=-10\), \(x = 1\)), D is at (1, -1) (on \(y=-1\), \(x = 1\)).

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

For D \((1,-1)\): \(-y=-(-1)=1\), \(-x=-1\), so \(D'=(1,-1)\)? Wait, no, if \(x = 1\), \(y=-1\), then \(-y = 1\), \(-x=-1\), so the new point is \((1,-1)\)? That seems odd. Wait, maybe the original D is (1, -1)? Wait, let's check the line \(y=-x\). The point (1, -1) is on \(y=-x\) (since \(-x=-1\) when \(x = 1\)), so reflecting a point on the line of reflection gives the same point. So D'=(1, -1) is correct.

For E \((-8,-1)\):
Substitute \(x=-8\) and \(y=-1\) into the rule \((x,y)\to(-y,-x)\).
\(-y=-(-1)=1\) and \(-x=-(-8)=8\). So \(E'=(1,8)\). Wait, no, wait: \((x,y)=(-8,-1)\), so \(-y = 1\), \(-x = 8\), so \(E'=(1,8)\)? Wait, no, the rule is \((x,y)\to(-y,-x)\). So \(x=-8\), \(y = - 1\), so \(-y=1\), \(-x = 8\), so the new point is \((1,8)\).

Wait, let's re - do each point:

1. Point B: Original coordinates \((x,y)=(-8,-10)\)

Reflection over \(y=-x\): \((x,y)\to(-y,-x)\), so \(-y=-(-10)=10\), \(-x=-(-8)=8\), so \(B'=(10,8)\)

1. Point C: Original coordinates \((x,y)=(1,-10)\)

Reflection over \(y=-x\): \(-y=-(-10)=10\), \(-x=-1\), so \(C'=(10,-1)\)

1. Point D: Original coordinates \((x,y)=(1,-1)\)

Reflection over \(y=-x\): \(-y=-(-1)=1\), \(-x=-1\), so \(D'=(1,-1)\) (since the point is on \(y = - x\), reflection leaves it unchanged)

1. Point E: Original coordinates \((x,y)=(-8,-1)\)

Reflection over \(y=-x\): \(-y=-(-1)=1\), \(-x=-(-8)=8\), so \(E'=(1,8)\)

Wait, maybe I made a mistake in original coordinates. Let's check the graph again. The purple figure: E is at (-8, -1) (x=-8, y=-1), B is at (-8, -10) (x=-8, y=-10), C is at (1, -10) (x = 1, y=-10), D is at (1, -1) (x = 1, y=-1). Yes.

So:

• \(B'=(10,8)\)
• \(C'=(10,-1)\)
• \(D'=(1,-1)\)
• \(E'=(1,8)\)

Answer:

\(B'(10, 8)\)
\(C'(10, -1)\)
\(D'(1, -1)\)
\(E'(1, 8)\)