Transform the equation$\text{ (ii) }3\mathrm{x}+4\mathrm{y}+12=0$

a) slope intercept form b) intercept form c) normal form

Given equation
$\begin{array}{l}3\mathrm{x}+4\mathrm{y}+12=0\Rightarrow 4\mathrm{y}=-3\mathrm{x}-12\\ \Rightarrow \mathrm{y}=\left(\frac{-3}{4}\right)\mathrm{x}-3\end{array}$
$\text{ Here }\mathrm{m}=\frac{-3}{4},\mathrm{c}=-3$
$\begin{array}{l}3\mathrm{x}+4\mathrm{y}+12=0\Rightarrow 3\mathrm{x}+4\mathrm{y}=-12\\ \Rightarrow \frac{3\mathrm{x}}{-12}+\frac{4\mathrm{y}}{-12}=1\\ \Rightarrow \frac{\mathrm{x}}{-4}+\frac{\mathrm{y}}{-3}=1\end{array}$
$$\begin{gathered} \text{ Normal from is }3\mathrm{x}+4\mathrm{y}=-12 \\ \Rightarrow -3\mathrm{x}-4\mathrm{y}=12 \\ dividing both sides by 5 \end{gathered}$$
$\begin{array}{l}\Rightarrow \left(\frac{-3}{5}\right)\mathrm{x}+\left(\frac{-4}{5}\right)\mathrm{y}=\frac{12}{5}\\ \Rightarrow \mathrm{xcos}\mathrm{\alpha }+\mathrm{ysin}\mathrm{\alpha }=\mathrm{p}\\ \cos \mathrm{\alpha }=\frac{-3}{5},\sin \mathrm{\alpha }=\frac{-4}{5},\mathrm{p}=\frac{12}{5}\end{array}$