If the three distinct lines x+2ay+a=0, x+3by+b=0 and x+4ay+a=0 are concurrent, then the po…

Question

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:

Accepted Answer

Subtracting the third from the first: $-2ay=0 \implies y=0$ (since $a
e0$ for distinct lines).

Substituting $y=0$ into the first: $x+a=0 \implies x=-a$.

Substituting into the second: $-a+0+b=0 \implies a=b$.

So $(a,b)$ satisfies $a=b$, i.e. lies on the straight line $y=x$.

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