TS ICET 23 May 2018 Shift 2 Paper

Question : 139 of 200

Marks: +1, -0

A point ( x,y) is equidistant from the points (7,1) and (3,5) then
(x - y) =

Solution:

Let P(x, y), A (7, 1), B(3, 5) be three points
Given PA = PB.

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

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

⇒-8 x+8 y=34-50

⇒-8(x-y)=-16

⇒x - = 2

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