Let's determine the correct position for point [tex]\( H \)[/tex] that ensures Alex uses no more than 20 units of fencing by considering the given points and the available fencing.
First, calculate the distances between the given points [tex]\( E \)[/tex], [tex]\( F \)[/tex], and [tex]\( G \)[/tex]:
1. Distance [tex]\( EF \)[/tex] between [tex]\( E(0, 5) \)[/tex] and [tex]\( F(5, 5) \)[/tex]:
[tex]\[
EF = \sqrt{(5 - 0)^2 + (5 - 5)^2} = \sqrt{5^2 + 0^2} = \sqrt{25} = 5 \text{ units}
\][/tex]
2. Distance [tex]\( FG \)[/tex] between [tex]\( F(5, 5) \)[/tex] and [tex]\( G(1, 1) \)[/tex]:
[tex]\[
FG = \sqrt{(1 - 5)^2 + (1 - 5)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \text{ units}
\][/tex]
3. Distance [tex]\( GE \)[/tex] between [tex]\( G(1, 1) \)[/tex] and [tex]\( E(0, 5) \)[/tex]:
[tex]\[
GE = \sqrt{(0 - 1)^2 + (5 - 1)^2} = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \text{ units}
\][/tex]
Next, calculate the total fencing used by summing these distances:
[tex]\[
\text{Total fencing used} = EF + FG + GE = 5 + 4\sqrt{2} + \sqrt{17}
\][/tex]
Given that Alex has 20 units of fencing, the remaining fencing available for the fourth side [tex]\( GH \)[/tex] is:
[tex]\[
\text{Remaining fencing} = 20 - (5 + 4\sqrt{2} + \sqrt{17})
\][/tex]
To decide where point [tex]\( H \)[/tex] could be placed, we evaluate the distances from point [tex]\( G \)[/tex] to each of the four potential points for [tex]\( H \)[/tex]:
1. Distance [tex]\( GH1 \)[/tex] between [tex]\( G (1, 1) \)[/tex] and [tex]\( H1 (-3, 1) \)[/tex]:
[tex]\[
GH1 = \sqrt{(-3 - 1)^2 + (1 - 1)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4 \text{ units}
\][/tex]
2. Distance [tex]\( GH2 \)[/tex] between [tex]\( G (1, 1) \)[/tex] and [tex]\( H2 (-3, -1) \)[/tex]:
[tex]\[
GH2 = \sqrt{(-3 - 1)^2 + (-1 - 1)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \text{ units}
\][/tex]
3. Distance [tex]\( GH3 \)[/tex] between [tex]\( G (1, 1) \)[/tex] and [tex]\( H3 (-5, 1) \)[/tex]:
[tex]\[
GH3 = \sqrt{(-5 - 1)^2 + (1 - 1)^2} = \sqrt{(-6)^2 + 0^2} = \sqrt{36} = 6 \text{ units}
\][/tex]
4. Distance [tex]\( GH4 \)[/tex] between [tex]\( G (1, 1) \)[/tex] and [tex]\( H4 (-5, -1) \)[/tex]:
[tex]\[
GH4 = \sqrt{(-5 - 1)^2 + (-1 - 1)^2} = \sqrt{(-6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \text{ units}
\][/tex]
Now, check which of these distances for points [tex]\( H1, H2, H3, H4 \)[/tex] satisfies the remaining fencing condition:
[tex]\[
4 \leq \text{remaining fencing}
\][/tex]
After checking the distances, it turns out that:
[tex]\[
GH1 = 4 \text{ units}
\][/tex]
Thus, point [tex]\( H \)[/tex] could be placed at [tex]\( (-3,1) \)[/tex] so that Alex does not have to buy more fencing. Therefore, the correct answer is:
[tex]\[
\boxed{(-3, 1)}
\][/tex]
