Question

Find the angle between the line $\sqrt{3}x+y+1=0$ and $x+1=0$.

Answer 1

When written in form $y=mx+c$, $m=$ slope of line

For $L_1:$ $y=-\sqrt{3}x-1=m_1=-\sqrt{3}$

For $L_2:o.y=x+1\Rightarrow y=\frac{x}{0}+\frac{1}{0}\Rightarrow m_2\to \infty$

Slope between lines $=\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|=\tan \theta$

$\theta =$Angle $\Rightarrow \theta ={\tan }^{-1}\left(\left|\frac{1-\frac{m_1}{m_2}}{m_1+\frac{1}{m_2}}\right|\right)={\tan }^{-1}\left(\left|\frac{1}{m_1}\right|\right)={\tan }^{-1}\left(\left|\frac{1}{\sqrt{3}}\right|\right)$

$\Rightarrow -\theta ={150}^o$ or ${30}^o$.