Transform the equation $\sqrt{3}\mathrm{x}+\mathrm{y}+10=0$
into (a) slope - intercept form (b) intercept form and (c) normalform.

$\text{ Given line }\sqrt{3}\mathrm{x}+\mathrm{y}+10=0$
(a) slope intercept form : (y = mx+c)
$\text{ given }\sqrt{3}\mathrm{x}+\mathrm{y}+10=0$
$\mathrm{y}=-\sqrt{3}\mathrm{x}-10\text{ where }\mathrm{m}=-\sqrt{3};\mathrm{c}=-10$
$\text{ (b) Intercept from : }\left(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\right)$
$$\begin{gathered} \text{ given }\sqrt{3}\mathrm{x}+\mathrm{y}+10=0 \\ \sqrt{3}\mathrm{x}+\mathrm{y}=-10 \\ dividing both sides by -10 \end{gathered}$$
$$\begin{gathered} \frac{\sqrt{3}\mathrm{x}}{-10}+\frac{\mathrm{y}}{-10}=\frac{-10}{-10} \\ \Rightarrow \frac{\mathrm{x}}{\frac{-10}{\sqrt{3}}}+\frac{\mathrm{y}}{-10}=1 \end{gathered}$$
$\mathrm{x}\text{- intercept }=\frac{-10}{\sqrt{3}};y\text{ - intercept }=-10$
$\text{ Normal form : }(\mathrm{xcos}\mathrm{\alpha }+\mathrm{ysin}\mathrm{\alpha }=\mathrm{P})$
$\text{ given }\sqrt{3}\mathrm{x}+\mathrm{y}+10=0$

dividing both sides by$\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=\sqrt{3+1}=\sqrt{4}=2$
$\begin{array}{l}\frac{\sqrt{3}}{2}+\frac{\mathrm{y}}{2}=\frac{-10}{2}\\ \frac{-\sqrt{3}}{2}\mathrm{x}-\frac{1}{2}\mathrm{y}=5 where\left(\mathrm{\theta }\in {\mathrm{Q}}_3\right)\\ x \cos \frac{7\mathrm{\pi }}{6}+y \sin \frac{7\mathrm{\pi }}{6}=5\end{array}$