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If lines represented by $x+3 y-6=0$, $2 x+y-4=0$ and $k x-3 y+1=0$ are concurrent, then the value of $k$ is
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Verified Answer
The correct answer is:
$\frac{19}{6}$
Given lines are,
$$
\begin{aligned}
&x+3 y-6=0 \\
&2 x+y-4=0 \\
&k x-3 y+1=0
\end{aligned}
$$
these given lines will concurrent
when
$$
\left|\begin{array}{rrr}
1 & 3 & -6 \\
2 & 1 & -4 \\
k & -3 & 1
\end{array}\right|=0
$$
Expand with respect to $R_{1}$ :
$1(1-12)-3(2+4 k)-6(-6-k)=0$
$\Rightarrow \quad-11-6-12 k+36+6 k=0$
$\Rightarrow \quad-6 k-17+36=0$
$\Rightarrow \quad 6 k=19$
$\Rightarrow \quad k=\frac{19}{6}$
$$
\begin{aligned}
&x+3 y-6=0 \\
&2 x+y-4=0 \\
&k x-3 y+1=0
\end{aligned}
$$
these given lines will concurrent
when
$$
\left|\begin{array}{rrr}
1 & 3 & -6 \\
2 & 1 & -4 \\
k & -3 & 1
\end{array}\right|=0
$$
Expand with respect to $R_{1}$ :
$1(1-12)-3(2+4 k)-6(-6-k)=0$
$\Rightarrow \quad-11-6-12 k+36+6 k=0$
$\Rightarrow \quad-6 k-17+36=0$
$\Rightarrow \quad 6 k=19$
$\Rightarrow \quad k=\frac{19}{6}$
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