To tackle this problem, we need to determine the coordinates of points [tex]\( A' \)[/tex] and [tex]\( P' \)[/tex] after the translation, given the original triangle with vertices [tex]\( Z(-1,5) \)[/tex], [tex]\( A(1,3) \)[/tex], and [tex]\( P(-2,4) \)[/tex]. We are informed that the translation maps the point [tex]\( Z(-1,5) \)[/tex] to [tex]\( Z'(1,1) \)[/tex].
First, we'll calculate the translation vector that maps [tex]\( Z \)[/tex] to [tex]\( Z' \)[/tex]. The translation vector can be found by subtracting the original coordinates of [tex]\( Z \)[/tex] from the coordinates of [tex]\( Z' \)[/tex]:
[tex]\[
\text{Translation vector} = (1 - (-1), 1 - 5) = (2, -4)
\][/tex]
With the translation vector [tex]\((2, -4)\)[/tex], we can now apply this translation to the other points [tex]\( A \)[/tex] and [tex]\( P \)[/tex].
### Translating Point [tex]\( A(1,3) \)[/tex]
To find [tex]\( A' \)[/tex], add the translation vector to the coordinates of [tex]\( A \)[/tex]:
[tex]\[
A' = (1 + 2, 3 - 4) = (3, -1)
\][/tex]
### Translating Point [tex]\( P(-2,4) \)[/tex]
Similarly, to find [tex]\( P' \)[/tex], add the translation vector to the coordinates of [tex]\( P \)[/tex]:
[tex]\[
P' = (-2 + 2, 4 - 4) = (0, 0)
\][/tex]
Thus, applying the translation vector [tex]\((2, -4)\)[/tex] to points [tex]\( A \)[/tex] and [tex]\( P \)[/tex] results in:
[tex]\[
A' = (3, -1) \quad \text{and} \quad P' = (0, 0)
\][/tex]
The coordinates of [tex]\( A' \)[/tex] and [tex]\( P' \)[/tex] under this translation are:
[tex]\[
\boxed{(3, -1) ; (0, 0)}
\][/tex]
Given the multiple-choice options, the correct answer is:
[tex]\[
A^{\prime}(3, -1) ; P(0, 0)
\][/tex]
