Transform the equation $4\mathrm{x}-3\mathrm{y}+12=0$
  a) slope intercept form b) intercept form  c) normal form

$\text{ Given line }4\mathrm{x}-3\mathrm{y}+12=0$
Slope intercept form is 
$\begin{array}{l}\mathrm{y}=\mathrm{mx}+\mathrm{c}\Rightarrow 3\mathrm{y}=4\mathrm{x}+12\\ \mathrm{y}=\frac{4}{3}\mathrm{x}+4\end{array}$
$$\begin{gathered} \text{ Intercept form is }\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1 \\ Given line 4x-3y+12=0 \\ \Rightarrow 4\mathrm{x}-3\mathrm{y}=-12 \\ dividing both sides by -12 \end{gathered}$$
$\frac{4\mathrm{x}}{-12}-\frac{3\mathrm{y}}{-12}=1\Rightarrow \frac{\mathrm{x}}{-3}+\frac{\mathrm{y}}{4}=1$
$$\begin{gathered} \text{ Normal form }\mathrm{xcos}\mathrm{\alpha }+\mathrm{ysin}\mathrm{\alpha }=\mathrm{p} \\ \mathrm{Given} \mathrm{line} 4\mathrm{x}-3\mathrm{y}+12=0 \\ \Rightarrow 4\mathrm{x}-3\mathrm{y}=-12 \\ \Rightarrow -4\mathrm{x}+3\mathrm{y}=12 \end{gathered}$$
$\text{ Dividing both sides by }\sqrt{(-4{)}^3+3^2}=5$
$\frac{-4\mathrm{x}}{5}+\frac{3\mathrm{y}}{5}=\frac{12}{5}$