Question

Find the angle between $\sqrt{3}x+y+16=0$ and $\sqrt{3}x-y+9=0$

Answer 1

We have,

$\sqrt{3}+y+16=0and\sqrt{3}-y+9=0$

then,

$\sqrt{3}+y+16=0---\left(1\right)$

$\sqrt{3}-y+9=0----\left(2\right)$

By equation (1)

$\sqrt{3}+y+16=0$

$y=-\sqrt{3}-16$

on compare that,

$y=m_{1}x+c$

$m_1=-\sqrt{3}$

By equation (2),

$\sqrt{3}-y+9=0$

$y=\sqrt{3}x+9$

on compare that,

$y=m_{2}x+c$

$m_2=\sqrt{3}$

then,

we know that angle between

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

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

$\tan \theta =\frac{2\sqrt{3}}{1+(\sqrt{3}{)}^2}$

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

$\tan \theta =\frac{2\sqrt{3}}{4}$

$\tan \theta =\frac{\sqrt{3}}{2}$

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

Hence, this is the answer.