Question

The angles between the line $\sqrt{3}x+y=1$ and $x-\sqrt{3}y=1$ is $\pi /x$ then $x=$

Answer 1

$\sqrt{3}x+y=1⟹m_1=-\sqrt{3}x+\sqrt{y}=1⟹m_2=\frac{1}{\sqrt{3}}$

Angle between them is

${\tan }^{-1}\frac{\frac{1}{\sqrt{3}}+\sqrt{3}}{1+\sqrt{3}(-\frac{1}{\sqrt{3}})}$

${\tan }^{-1}\infty =\frac{\pi }{2}$

$\therefore x=2$