Converting Linear Equations in Standard Form to Normal Form

The general equation or standard equation of a straight line is:
$ax + by + c = 0$

Where $a$ and $b$ are constants and either $a \ne 0$ or $b \ne 0$.

Convert the standard equation of line $ax + by + c = 0$ into the normal form $x\cos \alpha + y\sin \alpha = p$

The procedure of converting the equation into the normal form is as follows:
$\begin{gathered} ax + by + c = 0 \\ \Rightarrow – ax – by = c \\ \end{gathered}$

Dividing both sides of the equation by $\pm \sqrt {{a^2} + {b^2}}$  we have
$\Rightarrow – \frac{a}{{ \pm \sqrt {{a^2} + {b^2}} }}x – \frac{b}{{ \pm \sqrt {{a^2} + {b^2}} }}y = \frac{c}{{ \pm \sqrt {{a^2} + {b^2}} }}$

This is the equation of the line in normal form, where $\frac{c}{{ \pm \sqrt {{a^2} + {b^2}} }}$ is the length of the normal form origin of the line. Since the length of the normal form origin to the line must be positive, so:

(i) $– \frac{a}{{\sqrt {{a^2} + {b^2}} }}x – \frac{b}{{\sqrt {{a^2} + {b^2}} }}y = \frac{c}{{\sqrt {{a^2} + {b^2}} }}$ is the normal form of the line if $c > 0$.

(ii) $\frac{a}{{\sqrt {{a^2} + {b^2}} }}x + \frac{b}{{\sqrt {{a^2} + {b^2}} }}y = \frac{c}{{ – \sqrt {{a^2} + {b^2}} }}$ is the normal form of line if $c < 0$.

Example: Convert the equation $2x + 5y – 6 = 0$ into normal form.

The equation of a line in standard form is $2x + 5y – 6 = 0$
$\Rightarrow 2x + 5y = 6$

Dividing both sides of the equation by the normalizing factor $\sqrt {{{\left( 2 \right)}^2} + {{\left( 5 \right)}^2}} = \sqrt {29}$, we have
$\Rightarrow \frac{{2x}}{{\sqrt {29} }} + \frac{{5y}}{{\sqrt {29} }} = \frac{6}{{\sqrt {29} }}$

Compare with the normal form of a line $x\cos \alpha + y\sin \alpha = p$, where $\cos \alpha = \frac{2}{{\sqrt {29} }}$, $\sin \alpha = \frac{5}{{\sqrt {29} }}$and $p = \frac{6}{{\sqrt {29} }}$.