To solve this problem, we need to determine the translation vector used to move from trapezoid [tex]\( EFGH \)[/tex] to [tex]\( E'F'G'H' \)[/tex].
First, observe the translation of point [tex]\( E \)[/tex] to [tex]\( E' \)[/tex]:
- Point [tex]\( E \)[/tex] has coordinates [tex]\((-1, 4)\)[/tex].
- Point [tex]\( E' \)[/tex] has coordinates [tex]\((2, 2)\)[/tex].
To determine the translation vector that moves [tex]\( E \)[/tex] to [tex]\( E' \)[/tex]:
1. Calculate the change in the x-coordinate:
[tex]\[ \Delta x = E'_{\text{x}} - E_{\text{x}} = 2 - (-1) = 3 \][/tex]
2. Calculate the change in the y-coordinate:
[tex]\[ \Delta y = E'_{\text{y}} - E_{\text{y}} = 2 - 4 = -2 \][/tex]
Thus, the translation vector is [tex]\((\Delta x, \Delta y) = (3, -2)\)[/tex].
Next, we apply the same translation vector to point [tex]\( H \)[/tex]:
- Point [tex]\( H \)[/tex] has coordinates [tex]\((-3, 1)\)[/tex].
Applying the translation vector [tex]\((3, -2)\)[/tex] to [tex]\( H \)[/tex]:
[tex]\[
H'_{\text{x}} = H_{\text{x}} + 3 = -3 + 3 = 0
\][/tex]
[tex]\[
H'_{\text{y}} = H_{\text{y}} - 2 = 1 - 2 = -1
\][/tex]
Therefore, the coordinates of point [tex]\( H' \)[/tex] are [tex]\((0, -1)\)[/tex]. So, the correct answer is:
[tex]\[ H' = (0, -1) \][/tex]
