If the three distinct lines x + 2ay + a = 0,x | vedclass

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(C) The given lines are:$L_1: x + 2ay + a = 0$  $(1)$$L_2: x + 3by + b = 0$  $(2)$$L_3: x + 4ay + a = 0$  $(3)$Since the lines are concurrent,they int

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straight line

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(C) The given lines are:$L_1: x + 2ay + a = 0$  $(1)$$L_2: x + 3by + b = 0$  $(2)$$L_3: x + 4ay + a = 0$  $(3)$Since the lines are concurrent,they intersect at a common point.Subtracting equation $(1)$ from $(3)$:$(x + 4ay + a) - (x + 2ay + a) = 0$$2ay = 0$Since the lines are distinct,$a 
eq 0$,therefore $y = 0$.Substituting $y = 0$ into equation $(1)$:$x + 2a(0) + a = 0 \Rightarrow x = -a$.The point of concurrency is $(-a, 0)$.Since this point must lie on line $(2)$:$-a + 3b(0) + b = 0$$-a + b = 0 \Rightarrow b = a$.The point $(a, b)$ satisfies the equation $y = x$,which represents a straight line.

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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