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)

$\frac {19}{6}$

Answer 2

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