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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[/tex]-axis [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 :

GinnyAnswer
Sure! Let's determine the pre-image of vertex [tex]\( A' \)[/tex] using the given transformation rule [tex]\( r_y\text{-axe} (x, y) \rightarrow (-x, y) \)[/tex].

We have a list of points, which are the transformed vertices:

1. [tex]\( A' (4, 2) \)[/tex]
2. [tex]\( A' (2, -4) \)[/tex]
3. [tex]\( A' (-2, 4) \)[/tex]
4. [tex]\( A' (-4, -2) \)[/tex]

According to the given transformation rule [tex]\( (x, y) \rightarrow (-x, y) \)[/tex], we can find the corresponding pre-images by reversing the transformation:

1. For [tex]\( A' (4, 2) \)[/tex], the pre-image [tex]\((x, y)\)[/tex] is obtained by reversing the transformation:
[tex]\[ (-x, y) = (4, 2) \][/tex]
Thus, [tex]\( x = -4 \)[/tex], so the pre-image is:
[tex]\[ A (-4, 2) \][/tex]

2. For [tex]\( A' (2, -4) \)[/tex], the pre-image [tex]\((x, y)\)[/tex] is obtained by reversing the transformation:
[tex]\[ (-x, y) = (2, -4) \][/tex]
Thus, [tex]\( x = -2 \)[/tex], so the pre-image is:
[tex]\[ A (-2, -4) \][/tex]

3. For [tex]\( A' (-2, 4) \)[/tex], the pre-image [tex]\((x, y)\)[/tex] is obtained by reversing the transformation:
[tex]\[ (-x, y) = (-2, 4) \][/tex]
Thus, [tex]\( x = 2 \)[/tex], so the pre-image is:
[tex]\[ A (2, 4) \][/tex]

4. For [tex]\( A' (-4, -2) \)[/tex], the pre-image [tex]\((x, y)\)[/tex] is obtained by reversing the transformation:
[tex]\[ (-x, y) = (-4, -2) \][/tex]
Thus, [tex]\( x = 4 \)[/tex], so the pre-image is:
[tex]\[ A (4, -2) \][/tex]

Thus, the pre-images corresponding to the given transformed vertices [tex]\( A' \)[/tex] are:

- [tex]\( A (-4, 2) \)[/tex]
- [tex]\( A (-2, -4) \)[/tex]
- [tex]\( A (2, 4) \)[/tex]
- [tex]\( A (4, -2) \)[/tex]
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