Find the inclination of the line:

Solution:

(i) Given equation is $x+\sqrt{3} y+6=0$

We can rewrite it as $\sqrt{3} y=-x-6$

$\Rightarrow y=\frac{-1}{\sqrt{3}} x+\frac{-6}{\sqrt{3}}$

It is in the form of $\mathrm{y}=\mathrm{x} \times \tan \alpha+\mathrm{c}$

Where $\tan \alpha=-\frac{1}{\sqrt{3}}$ and $\mathrm{c}=-\frac{6}{\sqrt{3}}$

The inclination of the line is $\alpha$

Therefore $\alpha=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$

$=\frac{5 \pi}{6} 3 x+3 y=8$

Conclusion: Inclination $\mathrm{x}+\sqrt{3} \mathrm{y}+6=0$ of the line is $\frac{5 \pi}{6}$

$3 y=8-3 x$

(ii) Given equation is

We can rewrite it as

$\Rightarrow y=-x+\frac{-3}{8}$

It is in the form of $\mathrm{y}=\mathrm{x} \times \tan \alpha+\mathrm{c}$

Where $\tan \alpha=-1$ and $\mathrm{c}=-\frac{3}{8}$

Therefore $\alpha=\tan ^{-1}(-1)$

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

Conclusion: Inclination of line $3 x+3 y+8=0$ is $\frac{3 \pi}{4}$

(iii) Given equation is $\sqrt{3} \mathrm{x}-\mathrm{y}-4=0$

We can rewrite it as $\mathrm{y}=\sqrt{3} \mathrm{x}-4$

It is in the form of $\mathrm{y}=\mathrm{x} \times \tan \alpha+\mathrm{c}$

Where $\tan \alpha=\sqrt{3}$ and $c=-4$

$\Rightarrow \alpha=\tan ^{-1}(\sqrt{3})$

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

Conclusion: Inclination of the line is $\frac{\pi}{3}$