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

Question

Dr. Abubacker · Accepted Answer

(c)
Since, the given three lines are concurrent, then
$\left| \begin{matrix}
4 & 3 & - 1 \
1 & - 1 & 5 \
k & 5 & - 3
\end{matrix} ight| = 0$
$\Rightarrow \ 4(3 - 25) - 3( - 3 - 5k) - 1(5 + k) = 0$
$\Rightarrow \ \  - 88 + 9 + 15k - 5 - k = 0$
$\Rightarrow \ 14k = 84\  \Rightarrow \ k = 6$

Q. If the lines $4x + 3y - 1 = 0,\ x - y + 5 = 0$ and $kx + 5y - 3 = 0$ are concurrent, then $k$ is equal to

Explanation

Reviewed by Dr. Abubacker

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