Let $L_{1} : x + 3 y - 5 = 0$ , $L_{2} : 3 x - K y - 1 = 0$ , $L_{3} : 5 x + 2 y - 12 = 0$

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Solution

$L_{1} : x + 3 y - 5 = 0, L_{2} : 3 x - K y - 1 = 0 L_{3} : 5 x + 2 y - 12 = 0$
A) Solving $L_{1}, L_{3}$ we get $x = 2, y = 1$
Substituting this in $L_{2}$ we get $6 - K - 1 = 0 \Rightarrow K = 5$
B) For parralle line $m (L_{1}) = m (L_{2}) \Rightarrow - \frac{1}{3} = \frac{3}{K} \Rightarrow K = - 9$
And $m (L_{3}) = m (L_{2}) \Rightarrow - \frac{5}{2} = \frac{3}{K} \Rightarrow K = - \frac{6}{5}$
C) Three lines form a triangle when they are not concurrent and no two of them are parallel, therefore from (A) and (B) $K \neq 5, K \neq - 9$ and $K \neq - \frac{6}{5}$
D) From (A) ,(B) and (C)
$K = 5, K = - 9$ and $K = - \frac{6}{5}$

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Q. 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

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