The diagonals of the parallelogram whose sides are $l x + m y + n = 0, l x + m y + n_{1} = 0, m x + l y + n_{1} = 0$ include an angle

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$\frac{π}{3}$

$\frac{π}{2}$

$\tan^{- 1} [\frac{l^{2} - m^{2}}{l^{2} + m^{2}}]$

$\tan^{- 1} [\frac{2 l m}{l^{2} + m^{2}}]$

answer is B.

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Detailed Solution

Given lines

$\begin{array}{l} l x + m y + n = 0 - - (1) \\ l x + m y + n_{1} = 0 - - (2) \\ m x + l y + n = 0 - - (3) \\ m x + l y + n_{1} = 0 - - (4) \\ S l o p e o f (1) & (2) = - \frac{l}{m} \\ (1), (2) a r e p a r a l l e l d i s \tan c e b / w (1) & (2) = \frac{|n - n_{1}|}{\sqrt{l^{2} + m^{2}}} \end{array}$

$\begin{array}{l} S l o p e o f (3) & (4) = \frac{- m}{l} \\ ∴ (3), (4) a r e p a r a l l e l d i s \tan c e b / w (3) & (4) = \frac{|n - n_{1}|}{\sqrt{l^{2} + m^{2}}} \\ D i s \tan c e b / w (1) & (2) = D i s \tan c e b / w (3) & (4) a n d a d j c e n t s i d e s a r e n o t ⊥ r \\ ∴ T h e l i n e s f o r m e d R hom b u s \\ ∴ A n g l e b / w d i a g o n a l s = 90^{0} \end{array}$