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 :

Question

• circle
• hyperbola
• straight line
• parabola

Answer 1

x + 2ay + a = 0 (1) 

x + 3by + b = 0 (2) 

x + 4ay + a = 0  (3) 

As these lines are concurrent , so f we solve any two lines , then intersection point of these two lines will satisfy the third line. 

So take the line (1) and  (2), on solving we get

And pair of straight line equation is : px²+ qy² + 2hxy = 0 [ here p ,h and q are constant] 

So equation (1) also represent a pair of straight line. 

As the equation will be 4x²+ 6y²- 10xy  = 0  , put (a,b) .