Question

find a relation between x and y such that the points (x, y) is equidistant from the points (7, 1 ) and ( 3, 5 )

Answer 1

Heyy !! here is your answer

Let coordinates of point a=(x,y)
coordinates of point B =(7,1)
coordinates of C=(3,5)
distance between a and B = √(7-x)sq.+(1-y)sq.
=√49+(x)sq.-14x+1+(y)sq.-2 distance between a and C=√(3-x)sq.+(5-y)sq.
=√9+(x)sq.-6x+25+(y)sq.-10y


It is given that (x,y) is equidistant from point a and point B
So,
√49+(x)sq.-14x+1+(y)sq.-2 = √9+(x)sq.-6x+25+. (y)sq.-10y

root is present both sides so it is cancelled
Similarly (x)sq. and (y)sq. are cancelled


49-14x+1-2y=9-6x+25-10y
49+1-9-25=14x+2y-6x-10y
16=8x-10y
2(8)=2(4x-5y)
8=4x-5y
5y+8=4x. is the required relation