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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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Verified Answer
The correct answer is:
straight line
straight line
$$
\begin{aligned}
&x+2 a y+a=0 \\
&x+3 b y+b=0 \\
&x+4 a y+a=0
\end{aligned}
$$
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
\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
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