Question

which point would map onto itself after a reflection across the line $y = -x$?
$\bigcirc$ $(-4, -4)$
$\bigcirc$ $(-4, 0)$
$\bigcirc$ $(0, -4)$
$\bigcirc$ $(4, -4)$

Explanation:

Step1: Recall reflection over \( y = -x \)

The rule for reflecting a point \((x, y)\) over the line \( y=-x \) is \((x, y)\to(-y, -x)\). For a point to map onto itself, the image after reflection must be the same as the original point, so \((x, y)=(-y, -x)\). This implies \( x = -y \) and \( y = -x \), which simplifies to \( y=-x \). So the point must lie on the line \( y = -x \).

Step2: Check each option

• For \((-4, -4)\): Substitute into \( y=-x \). Left side \( y=-4 \), right side \( -x=-(-4)=4 \). \(-4

eq4\)? Wait, no, wait: Wait, the line is \( y = -x \), so for a point \((a,b)\) to be on \( y=-x \), \( b=-a \). Let's re - check:

• Option 1: \((-4, -4)\): If we apply the reflection rule \((x,y)\to(-y, -x)\), then \((-4,-4)\to(-(-4), -(-4))=(4,4)

eq(-4,-4)\). Wait, I made a mistake earlier. Wait, the correct condition for a point \((x,y)\) to be invariant under reflection over \( y = -x \) is that when we apply the reflection \((x,y)\to(-y, -x)\), we get \((x,y)\). So \( x=-y \) and \( y=-x \), which is the same as \( y=-x \). Wait, let's check each point:

• For \((-4, -4)\): Apply reflection: \((-(-4), -(-4))=(4,4)

eq(-4,-4)\).

• For \((-4,0)\): Apply reflection: \((-0, -(-4))=(0,4)

eq(-4,0)\).

• For \((0, -4)\): Apply reflection: \((-(-4), -0)=(4,0)

eq(0,-4)\).

• For \((4, -4)\): Apply reflection: \((-(-4), -4)=(4, -4)\). Wait, wait, let's do the reflection rule correctly. The reflection over \( y=-x \) is \((x,y)\to(-y, -x)\). So for \((4, -4)\): \( x = 4 \), \( y=-4 \). Then \(-y=-(-4)=4\), \(-x=-4\). So the image is \((4, -4)\), which is the same as the original point. Wait, also, let's check the line \( y=-x \): For \((4, -4)\), \( y=-4 \), \( -x=-4 \), so \( y=-x \) (since \(-4=-4\)). Ah, I see, earlier mistake in checking \((-4,-4)\): For \((-4,-4)\), \( y=-4 \), \( -x = 4 \), so \( y

eq -x \). For \((4, -4)\), \( y=-4 \), \( -x=-4 \), so \( y=-x \) holds. And when we reflect \((4, -4)\) over \( y=-x \), we get \((-(-4), -4)=(4, -4)\), which is the same as the original point.

Wait, I think I messed up the first check. Let's re - derive the invariant condition. If a point \((x,y)\) is invariant under reflection over \( y=-x \), then \((x,y)=(-y, -x)\). So we have the system of equations:
\( x=-y \)
\( y=-x \)

Substitute the second equation into the first: \( x=-(-x)=x \), which is an identity. So the condition is \( y = -x \). Now let's check each point:

• \((-4, -4)\): \( y=-4 \), \( -x = 4 \), \( -4

eq4 \), so not on \( y=-x \).

• \((-4,0)\): \( y = 0 \), \( -x=4 \), \( 0

eq4 \), not on \( y=-x \).

• \((0, -4)\): \( y=-4 \), \( -x = 0 \), \( -4

eq0 \), not on \( y=-x \).

• \((4, -4)\): \( y=-4 \), \( -x=-4 \), so \( y=-x \) holds. Now apply the reflection rule: \((x,y)\to(-y, -x)\), so \((4, -4)\to(-(-4), -4)=(4, -4)\), which is the same as the original point.

Answer:

\((4, -4)\) (assuming the last option is \((4, -4)\))