If the three distinct lines $x+2 a y+a=0, x+3 b y$ $+b=0$ and $x+4 a y+a=0$ are concurrent, then the point $(a, b)$ lies on $a$ :

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Aicharya · Accepted Answer

$ \begin{aligned} &x+2 a y+a=0 \ &x+3 b y+b=0 \ &x+4 a y+a=0 \end{aligned} $ Subtracting equation (3) from (1) $ \begin{aligned} &-2 a y=0 \ &a y=0 \Rightarrow y=0 \end{aligned} $ Putting value of $y$ in equation (1), we get $ \begin{aligned} &x+0+a=0 \ &x=-a \end{aligned} $ Putting value of $x$ and $y$ in equation (2), we get $ \begin{aligned} &-a+b=0 \ &a=b \end{aligned} $ Thus, $(a, b)$ lies on a straight line

If the three distinct lines $x+2 a y+a=0, x+3 b y$ $+b=0$ and $x+4 a y+a=0$ are concurrent, then the point…

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