Question

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

• A. ${\tan }^{-1}\left(\frac{17}{31}\right)$
• B. ${\tan }^{-1}\left(\frac{11}{35}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{7}\right)$
• D. ${\tan }^{-1}\left(\frac{33}{35}\right)$
• E. ${\tan }^{-1}\left(\frac{7}{33}\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{1}{7}\right)$

Given lines are $2x+11y-7=0$ and $x+3y+5=0$
$\Rightarrow y=-\frac{2}{11}x+\frac{7}{11}$
and $y=-\frac{1}{3}x-\frac{5}{3}$
Their, slopes are $m_1=-\frac{2}{11}$ and $m_2=-\frac{1}{3}$
Therefore, angle between lines is
$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$
$=\frac{-\frac{2}{11}+\frac{1}{3}}{1+\frac{2}{11}\times \frac{1}{3}}=\frac{5}{35}$
Thus $\theta ={\tan }^{-1}\left(\frac{1}{7}\right)$