If every point on the line (a1 – a2)x + (b1 – b2)y = c is equidistant from the points (a1, b1) and (a2, b2) then 2c =
a12−b12+a22−b22
a12+b12+a22+b22
a12+b12−a22−b22
none of these
Let (h, k) be any point on the given line then nh−a12+k−b12=h−a22+k−b22⇒ 2a1−a2h+2b1−b2k=a12+b12−a22−b22⇒ a1−a2h+b1−b2k =(1/2)a12+b12−a22−b22 (i)Since (h, k) lies on the given line a1−a2h+b1−b2k=c (ii)Comparing (i) and (ii) we get c=(1/2)a12+b12−a22−b22.