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A point has the coordinates [tex]\((m, 0)\)[/tex] where [tex]\(m \neq 0\)[/tex].

Which reflection of the point will produce an image located at [tex]\((0, -m)\)[/tex]?

A. A reflection of the point across the [tex]\(x\)[/tex]-axis

B. A reflection of the point across the [tex]\(y\)[/tex]-axis

C. A reflection of the point across the line [tex]\(y = x\)[/tex]

D. A reflection of the point across the line [tex]\(y = -x\)[/tex]

Sagot :

GinnyAnsweridn
Let's analyze the reflections for the point [tex]\( (m, 0) \)[/tex] where [tex]\( m \neq 0 \)[/tex].

1. Reflection across the [tex]\( x \)[/tex]-axis:
- When you reflect a point [tex]\( (x, y) \)[/tex] across the [tex]\( x \)[/tex]-axis, the [tex]\( y \)[/tex]-coordinate changes sign, while the [tex]\( x \)[/tex]-coordinate remains the same.
- Thus, reflecting [tex]\( (m, 0) \)[/tex] across the [tex]\( x \)[/tex]-axis yields [tex]\( (m, -0) = (m, 0) \)[/tex].

2. Reflection across the [tex]\( y \)[/tex]-axis:
- When you reflect a point [tex]\( (x, y) \)[/tex] across the [tex]\( y \)[/tex]-axis, the [tex]\( x \)[/tex]-coordinate changes sign, while the [tex]\( y \)[/tex]-coordinate remains the same.
- Thus, reflecting [tex]\( (m, 0) \)[/tex] across the [tex]\( y \)[/tex]-axis yields [tex]\( (-m, 0) \)[/tex].

3. Reflection across the line [tex]\( y = x \)[/tex]:
- When you reflect a point [tex]\( (x, y) \)[/tex] across the line [tex]\( y = x \)[/tex], the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] coordinates swap.
- Thus, reflecting [tex]\( (m, 0) \)[/tex] across the line [tex]\( y = x \)[/tex] yields [tex]\( (0, m) \)[/tex].

4. Reflection across the line [tex]\( y = -x \)[/tex]:
- When you reflect a point [tex]\( (x, y) \)[/tex] across the line [tex]\( y = -x \)[/tex], the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] coordinates swap and both change signs.
- Thus, reflecting [tex]\( (m, 0) \)[/tex] across the line [tex]\( y = -x \)[/tex] yields [tex]\( (0, -m) \)[/tex].

To determine the location of the image point [tex]\( (0, -m) \)[/tex], we can see from the analysis above that it is obtained by reflecting the original point [tex]\( (m, 0) \)[/tex] across the line [tex]\( y = -x \)[/tex].

Therefore, the correct reflection is across the line [tex]\( y = -x \)[/tex].
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