If the lines `ax + by + c = 0, bx + cy + a = 0` and `cx + ay + b = 0` be concurrent, then:

If the lines ax + ky + 10 =0, bx + (k+1) y + 10 =0 and cx + (k +2)y + 10=0 are concurrent, then

If the three distinct lines x + 2ay + a = 0 , x+ 3by + b = 0 and x + 4ay + a = 0 are concurrent , then the point (a,b) lies on a .

If the lines ax+by+c=0, bx+cy+a=0 and cx+ay+b=0 (a, b,c being distinct) are concurrent, then (A) a+b+c=0 (B) a+b+c=0 (C) ab+bc+ca=1 (D) ab+bc+ca=0

if a gt b gt c and the system of equations ax + by + cz = 0, bx + cy + az 0 and cx + ay + bz = 0 has a non-trivial solution, then the quadratic equation ax^(2) + bx + c =0 has

If the lines ax+by+c=0,bx+cy+a=0 and cx+ay+b=0a!=b!=c are concurrent then the point of concurrency is

If the lines (a-b-c) x + 2ay + 2a = 0 , 2bx + ( b- c - a) y + 2b = 0 and (2c+1) x + 2cy + c - a - b = 0 are concurrent , then the prove that either a+b+ c = 0 or (a+b+c)^(2) + 2a = 0

Solve: ax - by = 1 bx + ay = 0