Question

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

Answer 1

Let the slopes of the given lines be $m_1$ and $m_2$ respectively.

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

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

$\therefore m_1=-\sqrt{3}{m}_2=\frac{-1}{\sqrt{3}}$

$\text{tan}\theta$ = $\left|\begin{array}{c}\frac{m_2-m_1}{1+m_1{m}_2}\end{array}\right|$ = $\left|\begin{array}{c}\frac{\frac{-1}{\sqrt{3}}+\sqrt{3}}{\{1+(-\sqrt{3})\times \left(\frac{-1}{\sqrt{3}}\right)\}}\end{array}\right|$

= $\left|\begin{array}{c}\frac{2}{\sqrt{3}}\times \frac{1}{2}\end{array}\right|$ = $\left|\begin{array}{c}\frac{1}{\sqrt{3}}\end{array}\right|$

= $\frac{1}{\sqrt{3}}$ $\Rightarrow \theta ={30}^{\circ }({180}^{\circ }-\theta )=({180}^{\circ }-{30}^{\circ })={150}^{\circ }.$

Hence, the angles between the given lines are ${30}^{\circ }$ and ${150}^{\circ }$