Answer :
To determine which reflection of the point [tex]\((m, 0)\)[/tex] will produce an image located at [tex]\((0, -m)\)[/tex], let's consider the effects of reflecting the point across different axes and lines step by step.
1. Reflection across the [tex]\(x\)[/tex]-axis:
- Reflecting a point [tex]\((x, y)\)[/tex] across the [tex]\(x\)[/tex]-axis changes the sign of the [tex]\(y\)[/tex]-coordinate, resulting in [tex]\((x, -y)\)[/tex].
- Applying this to [tex]\((m, 0)\)[/tex] gives:
[tex]\[ (m, 0) \rightarrow (m, -0) = (m, 0) \][/tex]
- The image remains [tex]\((m, 0)\)[/tex], not [tex]\((0, -m)\)[/tex].
2. Reflection across the [tex]\(y\)[/tex]-axis:
- Reflecting a point [tex]\((x, y)\)[/tex] across the [tex]\(y\)[/tex]-axis changes the sign of the [tex]\(x\)[/tex]-coordinate, resulting in [tex]\((-x, y)\)[/tex].
- Applying this to [tex]\((m, 0)\)[/tex] gives:
[tex]\[ (m, 0) \rightarrow (-m, 0) \][/tex]
- The image becomes [tex]\((-m, 0)\)[/tex], not [tex]\((0, -m)\)[/tex].
3. Reflection across the line [tex]\(y = x\)[/tex]:
- Reflecting a point [tex]\((x, y)\)[/tex] across the line [tex]\(y = x\)[/tex] swaps the coordinates, resulting in [tex]\((y, x)\)[/tex].
- Applying this to [tex]\((m, 0)\)[/tex] gives:
[tex]\[ (m, 0) \rightarrow (0, m) \][/tex]
- The image becomes [tex]\((0, m)\)[/tex], not [tex]\((0, -m)\)[/tex].
4. Reflection across the line [tex]\(y = -x\)[/tex]:
- Reflecting a point [tex]\((x, y)\)[/tex] across the line [tex]\(y = -x\)[/tex] swaps the coordinates and changes the sign of both coordinates, resulting in [tex]\((-y, -x)\)[/tex].
- Applying this to [tex]\((m, 0)\)[/tex] gives:
[tex]\[ (m, 0) \rightarrow (0, -m) \][/tex]
- The image becomes [tex]\((0, -m)\)[/tex], which matches the desired image.
Therefore, the correct reflection that will produce an image located at [tex]\((0, -m)\)[/tex] is a reflection across the line [tex]\(y = -x\)[/tex].
The correct answer is:
[tex]\[ \text{A reflection of the point across the line } y = -x \][/tex]
1. Reflection across the [tex]\(x\)[/tex]-axis:
- Reflecting a point [tex]\((x, y)\)[/tex] across the [tex]\(x\)[/tex]-axis changes the sign of the [tex]\(y\)[/tex]-coordinate, resulting in [tex]\((x, -y)\)[/tex].
- Applying this to [tex]\((m, 0)\)[/tex] gives:
[tex]\[ (m, 0) \rightarrow (m, -0) = (m, 0) \][/tex]
- The image remains [tex]\((m, 0)\)[/tex], not [tex]\((0, -m)\)[/tex].
2. Reflection across the [tex]\(y\)[/tex]-axis:
- Reflecting a point [tex]\((x, y)\)[/tex] across the [tex]\(y\)[/tex]-axis changes the sign of the [tex]\(x\)[/tex]-coordinate, resulting in [tex]\((-x, y)\)[/tex].
- Applying this to [tex]\((m, 0)\)[/tex] gives:
[tex]\[ (m, 0) \rightarrow (-m, 0) \][/tex]
- The image becomes [tex]\((-m, 0)\)[/tex], not [tex]\((0, -m)\)[/tex].
3. Reflection across the line [tex]\(y = x\)[/tex]:
- Reflecting a point [tex]\((x, y)\)[/tex] across the line [tex]\(y = x\)[/tex] swaps the coordinates, resulting in [tex]\((y, x)\)[/tex].
- Applying this to [tex]\((m, 0)\)[/tex] gives:
[tex]\[ (m, 0) \rightarrow (0, m) \][/tex]
- The image becomes [tex]\((0, m)\)[/tex], not [tex]\((0, -m)\)[/tex].
4. Reflection across the line [tex]\(y = -x\)[/tex]:
- Reflecting a point [tex]\((x, y)\)[/tex] across the line [tex]\(y = -x\)[/tex] swaps the coordinates and changes the sign of both coordinates, resulting in [tex]\((-y, -x)\)[/tex].
- Applying this to [tex]\((m, 0)\)[/tex] gives:
[tex]\[ (m, 0) \rightarrow (0, -m) \][/tex]
- The image becomes [tex]\((0, -m)\)[/tex], which matches the desired image.
Therefore, the correct reflection that will produce an image located at [tex]\((0, -m)\)[/tex] is a reflection across the line [tex]\(y = -x\)[/tex].
The correct answer is:
[tex]\[ \text{A reflection of the point across the line } y = -x \][/tex]
