9514 1404 393
Answer:
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.
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* 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.
