Question

4. If the equation of the locus of a point equidistant from the points (a1, b1) and (a2, b2) is
(a1-a2) x + (b1-b2)y +c =0 then the value of c is​

Answer 1

Let (h,k) be the point on the locus. 

Then by the given conditions,

(h−a1)2+(k−b1)2=(h−a2)2+(k−b2)2

⇒2h(a1−a2)+2k(b1−b2)+a22−a12+b22−b12=0

⇒h(a1−a2)+k(b1−b2)+21(a22+b22−a12−b12)=0

Also, since (h,k) lies on the given locus, therefore

(a1−a2)x+(b1−b2)y+c=0

Comparing Eqs. (i) and (ii), we get

c= 1/2 (a2/2 +b2/2-a2/1-b2/1)