# 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 :

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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) .

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