If the lines $4 x+3 y-1=0, x-y+5=0$ and $k x+5 y-3=0$ are concurrent, then $k=$

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Aicharya · Accepted Answer

For concurrency $\left|\begin{array}{ccc}4 & 3 & -1 \ 1 & -1 & 5 \ k & 5 & -3\end{array}ight|=0$ $\Rightarrow 4(3-25)+3(5 k+3)-1(5+k)=0$ $\Rightarrow-88+15 k+9-5-k=0$ $\Rightarrow-84+14 k=0$ $\Rightarrow k=6$

If the lines $4 x+3 y-1=0, x-y+5=0$ and $k x+5 y-3=0$ are concurrent, then $k=$

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