Question

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

Answer 1

Consider the given equations,

$x-\sqrt{3}y=-1$ ….(1) and $\sqrt{3}x-y=-1$ ….(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,$\tan \theta =\frac{m_2-m_1}{1+m_1{m}_2}$

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

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

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

$\tan \theta =1$

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

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

Hence, this is the answer.