Consider the lines given by
$L_{1} : x + 3 y - 5 = 0$
$L_{2} : 3 x - k y - 1 = 0$
$L_{3} : 5 x + 2 y - 12 = 0$

Match the Statements / Expressions in List 1 with the Statements / Expressions in List 2 and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the $O R S .$

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Solution

$x + 3 y - 5 = 0$ and $5 x + 2 y - 12 = 0$ intersect at $(2, 1)$
Hence $6 - k - 1 = 0 k = 5$
For $L_{1}, L_{2}$ to be parallel.
$\frac{1}{3} = \frac{3}{- k} \Rightarrow k = - 9$
For $L_{2}, L_{3}$ to be parallel.
$\frac{3}{5} = \frac{- k}{2} \Rightarrow k = \frac{- 6}{5}$
For $k \neq 5, - 9, \frac{- 6}{5}$ they will form triangle.
For $k = 5, k = - 9, \frac{- 6}{5}$ they will not form triangle.

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