Question

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

Answer 1

Consider the given equations,

$x+\sqrt{3}y=1$..(1), $\sqrt{3}x+y=1$ ......(2)

from the equation 1^st,

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

Compare with, $y=m_{1}x+C$

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

From equation 2^nd ,

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

Compare with $y=m_{2}x+C$

$m_2=-\sqrt{3}$

Now,let the angle between line 1^st and 2^{nd $=\theta$},

Using the formula,$\tan \theta =\frac{m_2-m_1}{1+m_1{m}_2}$

$\tan \theta =\frac{-\sqrt{3}-(-\frac{1}{\sqrt{3}})}{1+(-\sqrt{3})\times (-\frac{1}{\sqrt{3}})}=\frac{-\sqrt{3}\times \sqrt{3}+1}{1+1}$

$\tan \theta =\frac{-3+1}{2}=-1$

$\tan \theta =-1$

$\tan \theta =-\tan \frac{\pi }{4}$

$\tan \theta =\tan (\pi -\frac{\pi }{4})$

$\tan \theta =\tan \frac{3\pi }{4}$

$\theta =\frac{3\pi }{4}$

Hence, this is the answer.