Question

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

Answer 1

We have $\sqrt{3}x$ + y = 1

$\Rightarrow y=-\sqrt{3}x+1$

$\therefore m_1=-\sqrt{3}$

Also $x+\sqrt{3}y=1$

$\Rightarrow \sqrt{3}y=-x+1$

$\Rightarrow y=\frac{-1}{3}x+\frac{1}{\sqrt{3}}$

$\therefore m_2=\frac{-1}{\sqrt{3}}$

Let $\theta$ be the angle between the lines. Then

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

$=\left|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}\right|=|\frac{-2}{\sqrt{3}}\times \frac{1}{2}|$

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

$tan\theta =tan{30}^{\circ }$ and $tan({180}^{\circ }-{30}^{\circ })$

$\Rightarrow$ $\theta ={30}^{\circ }$ and ${150}^{\circ }$