Acity planner designs a park that is a quadrilateral with vertices at j(−3, 1), k(1, 3), l(5, −1), and m(−1, −3). there is an entrance to the park at the midpoint of each side of the park. a straight path connects each entrance to the entrance on the opposite side. assuming each unit of the coordinate plane represents 10 meters, what is the total length of the paths to the nearest meter?

  104 m

Step-by-step explanation:

We can laboriously compute the positions of all of the midpoints, then find the difference of their coordinates, or we can see what those differences might be before we mess with any numbers.

The midpoint of JK is (J+K)/2; the midpoint of LM is (L+M)/2, so the difference in coordinates between these midpoints will be ...

  (J+K-L-M)/2 = ((K-M) +(J-L))/2 . . . . path from JK to LM .. (P1)

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Similarly, the midpoint of KL is (K+L)/2, and the midpoint of MJ is (M+J)/2. Then the difference between these midpoints is ...

  (K+L-M-J)/2 = ((K-M) -(J-L))/2 . . . . path from KL to MJ .. (P2)

So, the differences we want are ...

  K-M = (1, 3) -(-1, -3) = (2, 6)

  J-L = (-3, 1) -(5, -1) = (-8, 2)

And the differences in coordinates at the ends of the paths are ...

  P1 = ((2, 6) +(-8, 2))/2 = (-3, 4)

  P2 = ((2, 6) -(-8, 2))/2 = (5, 2)

The total path length is found by applying the Pythagorean theorem (distance formula) to the differences in coordinates ...

  |P1| +|P2| = (10 m)×(√((-3)²+4²) +√(5² +2²)) = (10 m)(√25 +√29)

  |P1| +|P2| ≈ (10 m)(10.385) ≈ 103.85 m ≈ 104 m

The total path length is about 104 meters.

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In the attached graph, the lengths of the cross-park paths in coordinate units are shown at lower left.

answer: yes i agree that you should not put it on/p>

step-by-step explanation: the reason for this is that it wastes other peoples time and also make them fail as well and that is never good.