Question

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

Answer 1

The given lines are $\sqrt{3}x+y=1$ and $x+\sqrt{3}y=1$
$\Rightarrow y=-\sqrt{3}x+1$ and $y=-\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}$
Thus slope of line (1) is $m_1=-\sqrt{3}$ while the slope of line (2) is $m_2=-\frac{1}{\sqrt{3}}$
The acute angle i.e;$\theta$ between the two lines is given by
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$=|-\frac{\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3})(-\frac{1}{\sqrt{3}})}|$
$=\left|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}\right|=\left|\frac{-2}{2\sqrt{3}}\right|$
$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}},\theta ={30}^{\circ }$
Thus, the angle between the given lines is either ${30}^{\circ }$ or ${180}^{\circ }-{30}^{\circ }={150}^{\circ }$