Question

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

Answer 1

$L_1:\sqrt{3}x+y=1;L_2:x+\sqrt{3}y=1$

$m_1=-\sqrt{3};m_2=\frac{1}{\sqrt{3}}$

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

$=\left|\frac{-\sqrt{3}-\frac{1}{\sqrt{3}}}{1+(-\sqrt{3})\left(\frac{-1}{\sqrt{3}}\right)}\right|$

$=\left|\frac{\frac{-3-1}{\sqrt{3}}}{1+1}\right|$

$=\frac{\frac{-4}{\sqrt{3}}}{\frac{2}{1}}$

$=\frac{-2}{\sqrt{3}}$

$\therefore \tan \theta =\frac{-2}{\sqrt{3}}$

$\therefore \theta ={\tan }^{-1}\left(\frac{-2}{\sqrt{3}}\right)$.