Answer :
To determine which reflection of the point \((m, 0)\) will produce an image located at \((0, -m)\), we need to analyze the effects of reflecting a point across different lines.
1. Reflection across the \(x\)-axis:
- Reflecting a point \((a, b)\) across the \(x\)-axis results in the point \((a, -b)\).
- For the point \((m, 0)\), reflecting across the \(x\)-axis would give us \((m, 0)\), which is the same point. This does not match the image \((0, -m)\).
2. Reflection across the \(y\)-axis:
- Reflecting a point \((a, b)\) across the \(y\)-axis results in the point \((-a, b)\).
- For the point \((m, 0)\), reflecting across the \(y\)-axis would give us \((-m, 0)\). This result is also not \((0, -m)\).
3. Reflection across the line \(y = x\):
- Reflecting a point \((a, b)\) across the line \(y = x\) results in the point \((b, a)\).
- For the point \((m, 0)\), reflecting across the line \(y = x\) would give us the point \((0, m)\). This does not match our target image \((0, -m)\).
4. Reflection across the line \(y = -x\):
- Reflecting a point \((a, b)\) across the line \(y = -x\) results in the point \((-b, -a)\).
- For the point \((m, 0)\), reflecting across the line \(y = -x\) would transform it to \((0, -m)\). This is the correct image we are looking for.
Thus, the reflection of the point \((m, 0)\) that produces the image \((0, -m)\) is across the line \(y = -x\).
Therefore, the correct answer is:
- A reflection of the point across the line [tex]\(y = -x\)[/tex].
1. Reflection across the \(x\)-axis:
- Reflecting a point \((a, b)\) across the \(x\)-axis results in the point \((a, -b)\).
- For the point \((m, 0)\), reflecting across the \(x\)-axis would give us \((m, 0)\), which is the same point. This does not match the image \((0, -m)\).
2. Reflection across the \(y\)-axis:
- Reflecting a point \((a, b)\) across the \(y\)-axis results in the point \((-a, b)\).
- For the point \((m, 0)\), reflecting across the \(y\)-axis would give us \((-m, 0)\). This result is also not \((0, -m)\).
3. Reflection across the line \(y = x\):
- Reflecting a point \((a, b)\) across the line \(y = x\) results in the point \((b, a)\).
- For the point \((m, 0)\), reflecting across the line \(y = x\) would give us the point \((0, m)\). This does not match our target image \((0, -m)\).
4. Reflection across the line \(y = -x\):
- Reflecting a point \((a, b)\) across the line \(y = -x\) results in the point \((-b, -a)\).
- For the point \((m, 0)\), reflecting across the line \(y = -x\) would transform it to \((0, -m)\). This is the correct image we are looking for.
Thus, the reflection of the point \((m, 0)\) that produces the image \((0, -m)\) is across the line \(y = -x\).
Therefore, the correct answer is:
- A reflection of the point across the line [tex]\(y = -x\)[/tex].
