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Arada [10]
3 years ago
8

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

nd G (6, 1). He has 16 units of fencing. Where could Alex place point H so that he does not have to buy more fencing?
answer choices
(0,1)
(0,-2)
(1,1)
(1,-2)
Mathematics
1 answer:
hram777 [196]3 years ago
7 0

Answer:

(1,1)

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points (x_1,y_1)\,,\,(x_2,y_2) is equal to \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

So,

EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}

Perimeter of a figure is the length of its outline.

EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

Put (x,y)=(1,1)

\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9

This is true.

So, the point (1,1) satisfies the equation \sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

So, point H is (1,1).

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