Question

If 4a^2+9b^2-c^2+12ab=0 then family of straight lines ax+by+c=0 is concurrent is

Answer 1

Let  a1 = a/c   and   b1 = b/c

Then   4 a1² + 9 b1² - 1 + 12 a1 b1 = 0 
       =>  (2 a1 + 3 b1)² = 1
       =>   2 a1 + 3 b1  = +1  or  -1
       =>   a1 = (1 - 3 b1)/2   or  - (1 + 3 b1)/2

  and       a1 x + b1 y + 1 = 0  
   => (1 - 3b1) x +2 b1 y +2 = 0        or,   -(1+3 b1) x +2 b1 y+2 =0
   =>  (x + 2) + b1 (2 y - 3x) = 0        or    (2-x)  + b1 (2y - 3x) = 0
   =>  concurrent lines passing through (-2, -3)    or   (2, 3)