Question

If the point (x, y) is equidistant from the point (7,1) and (3,5) then find a relation between x and y

Answer 1

HEY MATE!

HERE IS THE STEP BY STEP EXPLANATION:

If p(x,y) is equidistant from point A(7,1) and B(3,5)

so, distance of AP=BP

i.e root (x-7)^2+(y-1)^2=root(x-3)^2+(y-5)^2

Squaring both sides we get,

=(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2

=x^2+49-14x+y^2+1-2y=x^2+9-6x+y^2+25-10y

=49-14x+1-2y=9-6x+26-10y

=50-14x-2y=35-6x-10y

=50-35-14x+6x-2y+10y=0

=15-8x+8y=0

=8y-8x=-15

=y-x=-15/8

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HOPE IT HELPED YOU ^_^