Transform the equation
$\text{ (i) }\sqrt{3}\mathrm{x}+\mathrm{y}=4$  into a)slope intercept form b)intercept form c)normal form

$\text{ i) Given equation }\sqrt{3}\mathrm{x}+\mathrm{y}=4$
$\text{ a) Slope intercept form }\sqrt{3}\mathrm{x}+\mathrm{y}=4$
$\mathrm{y}=-\sqrt{3}\mathrm{x}+4$
which is in the form of y = mx + c
$\therefore \text{ slope }\left(\mathrm{m}\right)=-\sqrt{3},\mathrm{y}\text{-intercept }=4$
$$\begin{gathered} \text{ b) Intercept form }\sqrt{3}\mathrm{x}+\mathrm{y}=4 \\ dividing both sides by 4 \end{gathered}$$
$$\begin{gathered} \frac{\sqrt{3}}{4}\mathrm{x}+\frac{1}{4}\mathrm{y}=1 \\ \Rightarrow \frac{\mathrm{x}}{{\displaystyle \frac{4}{\sqrt{3}}}}+\frac{\mathrm{y}}{4}=1 \end{gathered}$$
$\text{ which is in the form of }\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$
$\mathrm{x}\text{-intercept }=\frac{4}{\sqrt{3}};\mathrm{y}\text{-intercept }=4$
$\text{ c) Normal form : }\sqrt{3}\mathrm{x}+\mathrm{y}=4$
$\text{ Dividing with }\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=\sqrt{3+1}=2\text{ o.b.s }$
$\begin{array}{l}\Rightarrow \mathrm{x}\left(\frac{\sqrt{3}}{2}\right)+\mathrm{y}\left(\frac{1}{2}\right)=2\\ \therefore \cos \mathrm{\alpha }=\frac{\sqrt{3}}{2};\sin \mathrm{\alpha }=\frac{1}{2}\\ \therefore \mathrm{\alpha }={30}^{\circ }\text{ and }\mathrm{p}=2\end{array}$
The normal form is $\mathrm{xcos}\frac{\mathrm{\pi }}{6}+\mathrm{ysin}\frac{\mathrm{\pi }}{6}=2$