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

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

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

$\tan \theta =1$

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

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

Hence, this is the answer.