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What is the pre-image of vertex [tex]A^{\prime}[/tex] if the rule that created the image is [tex]r_{y \text{-axis}}(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 each vertex when the transformation rule is [tex]\( r_{y\text{-axis}}(x, y) \rightarrow (-x, y) \)[/tex], we need to reflect each given point across the y-axis. The transformation rule indicates that the x-coordinate of the point is negated, while the y-coordinate remains unchanged.

Here's the step-by-step reflection process for each given vertex:

1. For [tex]\( A'(-4, 2) \)[/tex]:
- The x-coordinate is [tex]\( -4 \)[/tex]. To find the pre-image, we negate the x-coordinate: [tex]\( -(-4) = 4 \)[/tex].
- The y-coordinate is [tex]\( 2 \)[/tex] and remains unchanged.
- Thus, the pre-image of [tex]\( A'(-4, 2) \)[/tex] is [tex]\( A(4, 2) \)[/tex].

2. For [tex]\( A'(-2, -4) \)[/tex]:
- The x-coordinate is [tex]\( -2 \)[/tex]. To find the pre-image, we negate the x-coordinate: [tex]\( -(-2) = 2 \)[/tex].
- The y-coordinate is [tex]\( -4 \)[/tex] and remains unchanged.
- Thus, the pre-image of [tex]\( A'(-2, -4) \)[/tex] is [tex]\( A(2, -4) \)[/tex].

3. For [tex]\( A'(2, 4) \)[/tex]:
- The x-coordinate is [tex]\( 2 \)[/tex]. To find the pre-image, we negate the x-coordinate: [tex]\( -(2) = -2 \)[/tex].
- The y-coordinate is [tex]\( 4 \)[/tex] and remains unchanged.
- Thus, the pre-image of [tex]\( A'(2, 4) \)[/tex] is [tex]\( A(-2, 4) \)[/tex].

4. For [tex]\( A'(4, -2) \)[/tex]:
- The x-coordinate is [tex]\( 4 \)[/tex]. To find the pre-image, we negate the x-coordinate: [tex]\( -(4) = -4 \)[/tex].
- The y-coordinate is [tex]\( -2 \)[/tex] and remains unchanged.
- Thus, the pre-image of [tex]\( A'(4, -2) \)[/tex] is [tex]\( A(-4, -2) \)[/tex].

Combining these results, the pre-images of the given vertices are:
[tex]\[ ((4, 2), (2, -4), (-2, 4), (-4, -2)) \][/tex]