If the lines $L_{1} : 2 x + y - 3 = 0$ , $L_{2} : 5 x + k y - 3 = 0$ and $L_{3} : 3 x - y - 2 = 0$ , are concurrent, then the value of $k$ is

- 2

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5

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- 3

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3

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Solution

The correct option is A $- 2$
Since lines are concurrent
$∣ ∣ ∣ ∣ \begin{matrix} 2 & 1 & - 3 5 & k & - 3 3 & - 1 & - 2 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$∴ 2 (- 2 k - 3) - 1 (- 10 + 3) - 3 (- 5 - 3 k) = 0$
$∴ - 9 k - 6 + 1 + 15 + 3 k = 0$
$∴ 10 + 5 k = 0$
$∴ k = \frac{- 10}{5}$
$∴ k = - 2$

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Similar questions

2 x + y - 3 = 0, 5 x + k y - 3 = 0

3 x - y - 2 = 0

If the lines 2x + y- 3 = 0, 5x + ky – 3 = 0 and 3x – y – 2 = 0 are concurrent,then the value of k is

L_{1} = 3 x + y - 5 = 0, L_{2} = 2 x + y - 3 = 0, L_{3} = 3 x + 2 y - 7 = 0

k x + 2 y + 2 = 0,

2 x + k y + 3 = 0,

3 x + 3 y + k = 0

L_{1} : x + 3 y - 5 = 0

L_{2} : 3 x - k y - 1 = 0

L_{3} : 5 x + 2 y - 12 = 0

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