Question

A city 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?

Answer 1

The Quadrilateral is JKLM,

let
`M_(JK), M_(KL), M_(LM), M_(JM),` be the midpoints of JK, KL, LM and JM respectively.

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Given any 2 point P(m,n) and Q(k,l),

the coordinates of the midpoint of the line segment PQ are given by the formula:


`M_(PQ)=( (m+k)/(2) , (n+l)/(2))`,

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thus the coordinates of points
`M_(JK), M_(KL), M_(LM), M_(JM),`

are as follows:

`M_(JK)= ((-3+1)/(2), (1+3)/(2))=(-1,2), \\\\M_(KL)= ((1+5)/(2), (3-1)/(2)=(3,1), \\\\M_(LM)= ((5-1)/(2), (-1-3)/(2))=(2, -2),\\\\ M_(JM)= ((-3-1)/(2), (1-3)/(2))=(-2,-1)`

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The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:


`|PQ|= \sqrt{ (a-c)^(2) + (b-d)^(2) }`

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thus the distances connecting the opposite entrances can be calculated as follows:

`|M_(JK),M_(LM)|= \sqrt{ (-1-2)^(2) + (2-(-2))^(2) }= √(9+16)=5`

`|M_(KL)M_(JM)|= \sqrt{ (3-(-2))^(2) + (1-(-1))^(2)}= √(25+4)= √(29)=5.39`

Thus the total distance of the paths joining the opposite entrances is

5+5.39 units = 50 m + 53.9 m = 104 m (rounded to the nearest meter)

Answer: 104 m