EASY 100 POINTS WILL MARK BRAINIEST Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (3,5), F (6,5), and G (9, 1). He has 22 units of fencing.
Where could Alex place point H so that he does not have to buy more fencing?
(0,0)
O(-1,0)
(0, -3)
(0,3)

D - (0,3)

Step-by-step explanation:

The length of fence already used is ...

 EF +FG = 3 + √((9-6)² +(5-1)²) = 3 +5 = 8

So, the remaining segments of fence must total less than 22 -8 = 14 units.

A plot of the points shows you that D(0, 3) is closest to the other three. An estimate* of the distance from the next-closest point, A(0, 0), shows it must be greater than 5+9 = 14, so only D is a possible location.

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In the attached we have shown an ellipse that is the set of all points such that their total distances to E and G is 14. Any point that is a solution to the problem must lie on or in that ellipse.

* For the purpose of estimating the distance between points, you can always use the maximum of the difference in x-coordinates or the difference in y-coordinates. The actual distance will be at least this value.

EA has an x-difference of 3 and a y-difference of 5, so the length EA is more than 5. GA has an x-difference of 9 and a y-difference of 1, so GA is longer than 9. That means the total EA +GA > 5+9 = 14. In short, point A cannot be a solution to the problem. Points B and C are even farther away.