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What is the pre-image of vertex [tex]\( A' \)[/tex] if the rule that created the image is [tex]\( r_{y\text{-axis}} \)[/tex]: [tex]\( (x, y) \rightarrow (-x, y) \)[/tex]?

A. [tex]\( A(-4, 2) \)[/tex]
B. [tex]\( A(-2, -4) \)[/tex]
C. [tex]\( A(2, 4) \)[/tex]
D. [tex]\( A(4, -2) \)[/tex]


Sagot :

To find the pre-image of vertex [tex]\( A' \)[/tex] given the reflection rule across the y-axis [tex]\( (x, y) \rightarrow (-x, y) \)[/tex], we need to reverse this transformation.

The inverse of the rule [tex]\( (-x, y) \)[/tex] is [tex]\( (x, y) \)[/tex]. This means if we have a point [tex]\( A' \)[/tex], its coordinates [tex]\((-x', y')\)[/tex] must be transformed back into [tex]\((x, y)\)[/tex].

We are given the point [tex]\( A' \)[/tex] which we will denote as [tex]\( (4, -2) \)[/tex].

1. According to the reflection rule [tex]\( (x, y) \rightarrow (-x, y) \)[/tex], the coordinates of [tex]\( A' \)[/tex] after reflection are [tex]\( (4, -2) \)[/tex].

2. To find the pre-image, we need to determine the original point [tex]\( A \)[/tex] before the transformation. Given the coordinates of [tex]\( A' \)[/tex] as [tex]\( (4, -2) \)[/tex]:
- The [tex]\( x \)[/tex]-coordinate of [tex]\( A \)[/tex] must have been the opposite sign of the [tex]\( x \)[/tex]-coordinate of [tex]\( A' \)[/tex].
- The [tex]\( y \)[/tex]-coordinate of [tex]\( A \)[/tex] remains the same as the [tex]\( y \)[/tex]-coordinate of [tex]\( A' \)[/tex].

3. Therefore, if [tex]\( A' = (4, -2) \)[/tex], to find [tex]\( A \)[/tex]:
- The [tex]\( x \)[/tex]-coordinate changes from [tex]\( 4 \)[/tex] to [tex]\( -4 \)[/tex]
- The [tex]\( y \)[/tex]-coordinate remains [tex]\( -2 \)[/tex]

Thus, the pre-image [tex]\( A \)[/tex] is [tex]\((-4, -2)\)[/tex].

Among the choices provided:
- [tex]\( A(-4, 2) \)[/tex]
- [tex]\( A(-2, -4) \)[/tex]
- [tex]\( A(2, 4) \)[/tex]
- [tex]\( A(4, -2) \)[/tex]

The correct pre-image is [tex]\( A(-4, -2) \)[/tex].
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