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 : px^{2}+ qy^{2} + 2hxy = 0 [ here p ,h and q are constant]

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

As the equation will be 4x^{2}+ 6y^{2}- 10xy = 0 , put (a,b) .