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

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Solution

(a) $L_{1}$ and $L_{3}$ intersect at $(1, 2)$ and it will lie on $L_{2}$ if $k = 5$
(b) Condition of parallelism gives
$\frac{3}{1} = - \frac{k}{3} = \frac{1}{5}$ $∴ k = - 9$

$∴ (b) \to (p)$
Also $\frac{3}{3} = - \frac{k}{2} = \frac{1}{12}$ $∴ k = - \frac{6}{5}$ $∴ (b) \to (q)$

$∴ (b) \to (p, q)$
(c) If $k \neq 5, - 9, - \frac{6}{5}$ i.e. (a) and (b) are ruled out and hence they form a triangle.

$∴ (c) \to (r)$
(d) Exclude the cases of (a) and (b) i.e.,

$k = 5, - 9, - \frac{6}{5}$ , they will not form a triangle.

$∴ (d) \to (p, q, s)$

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Q. Let

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

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

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

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/Exp[ressions in Column I with the Statements/Expressions in Column II.
$\begin{matrix} Column I & Column II L_{1}, L_{2}, L_{3} are concurrent, if & k = -9 One of L_{1}, L_{2}, L_{3} is parallel to at least one of the other two, if & k = \frac{6}{5} L_{1}, L_{2}, L_{3} form a triangle, if & k = \frac{5}{6} L_{1}, L_{2}, L_{3} do not form a triangle, if & k = 5 \end{matrix}$