The angle between the lines 2x + 11y - 7 = 0 | vedclass

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(C) The given lines are $2x + 11y - 7 = 0$ and $x + 3y + 5 = 0$. The slopes of these lines are $m_1 = -\frac{2}{11}$ and $m_2 = -\frac{1}{3}$. The ang

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$	an^{-1}\left(\frac{1}{7}ight)$

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(C) The given lines are $2x + 11y - 7 = 0$ and $x + 3y + 5 = 0$. The slopes of these lines are $m_1 = -\frac{2}{11}$ and $m_2 = -\frac{1}{3}$. The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}ight|$. Substituting the values: $	an 	heta = \left|\frac{-\frac{2}{11} - (-\frac{1}{3})}{1 + (-\frac{2}{11})(-\frac{1}{3})}ight| = \left|\frac{-\frac{6}{33} + \frac{11}{33}}{1 + \frac{2}{33}}ight| = \left|\frac{\frac{5}{33}}{\frac{35}{33}}ight| = \frac{5}{35} = \frac{1}{7}$. Therefore,$	heta = 	an^{-1}\left(\frac{1}{7}ight)$.

The angle between the lines $2x + 11y - 7 = 0$ and $x + 3y + 5 = 0$ is

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