Question

If lines represented by $x+3y-6=0,2x+y-4=0$ and $kx-3y+1=0$ are concurrent, then the value of $k$ is

• A. $\frac{6}{19}$
• B. $\frac{19}{6}$
• C. $\frac{-19}{6}$
• D. $\frac{-6}{19}$

Answer 1

Answer: (B)

Given lines are,
$x+3y-6=0$
$2x+y-4=0$
$kx-3y+1=0$
these given lines will concurrent
when $\left|\begin{array}{ccc}1& 3& -6\\ 2& 1& -4\\ k& -3& 1\end{array}\right|=0$
Expand with respect to $R_1$ :
$1(1-12)-3(2+4k)-6(-6-k)=0$
$\Rightarrow -11-6-12k+36+6k=0$
$\Rightarrow -6k-17+36=0$
$\Rightarrow 6k=19$
$\Rightarrow k=\frac{19}{6}$