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If the lines $4 x+3 y-1=0, x-y+5=0$ and $k x+5 y-3=0$ are concurrent, then $k$ is equal to :
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The correct answer is:
6
Since, the given three lines are concurrent, then
$\left|\begin{array}{rrr}4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3\end{array}\right|=0$
$\Rightarrow 4(3-25)-3(-3-5 k)-1(5+k)=0$
$\Rightarrow-88+9+15 k-5-k=0$
$\Rightarrow \quad 14 k=84$
$\therefore \quad k=6$
$\left|\begin{array}{rrr}4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3\end{array}\right|=0$
$\Rightarrow 4(3-25)-3(-3-5 k)-1(5+k)=0$
$\Rightarrow-88+9+15 k-5-k=0$
$\Rightarrow \quad 14 k=84$
$\therefore \quad k=6$
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