If lines $2 x + y - 3 = 0, 5 x + k y - 3 = 0$ and $3 x - y - 2 = 0$ are concurrent, then value of $k$ will be

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Solution

The correct option is B $- 2$
$\frac{x}{- 3 + 3 k} = \frac{y}{- 15 + 6} = \frac{1}{2 k - 5}$
$\Rightarrow x = \frac{- 3 + 3 k}{2 k - 5}, y = \frac{- 9}{2 k - 5}$
Substitute the values of $x$ and $y$ in $3 x - y - 2 = 0$
$3 (\frac{- 3 + 3 k}{2 k - 5}) - (\frac{- 9}{2 k - 5}) - 2 = 0$
$\Rightarrow - 9 + 9 k + 9 - 4 k + 10 = 0$
$\Rightarrow 5 k + 10 = 0$
$∴ k = \frac{- 10}{5} = - 2$

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