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(Q23)Find the relation between x and y such that the point (x,y) is equidistant from the points(7,1) and (3,5).

Answer:

Given P(x,y) be equidistant from the points A(7,1) and B(3,5)

AP=BP. So, AP²=BP²

(x-7)²+(y-1)²=(x-3)²+(y-5)²

(x²-14x+)+(y²-2y+)= (x²-6x+)+(y²-10y+)

(x²+y²-14x-2y+) - (x²+y²-6x-10y+)=0

-8x+8y= -

x-y= which is the required relation.