Answer:
[tex](1,1)[/tex]
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 [tex](x_1,y_1)\,,\,(x_2,y_2)[/tex] is equal to [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
So,
[tex]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}[/tex]
Perimeter of a figure is the length of its outline.
[tex]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[/tex]
Put [tex](x,y)=(1,1)[/tex]
[tex]\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[/tex]
This is true.
So, the point [tex](1,1)[/tex] satisfies the equation [tex]\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9[/tex]
So, point H is [tex](1,1)[/tex].
