Different Forms of $\displaystyle Ax+By+C=0$
Slope-intercept Form
If $\displaystyle B\neq 0$, then $\displaystyle Ax+By+C=0$ can be written as
$\displaystyle y=-\frac{A}{B}x-\frac{C}{B}$
$\displaystyle y=mx+c$
where, slope $\displaystyle m=-\frac{A}{B}$
y-intercept $\displaystyle c=-\frac{C}{B}$
If $\displaystyle B=0$, then slope is undefined
x-intercept is $\displaystyle -\frac{C}{A}$
Intercept Form
If $\displaystyle B\neq 0$, then $\displaystyle Ax+By+C=0$ can be written as
$\displaystyle \frac{x}{-\frac{C}{A}}+\frac{y}{-\frac{C}{B}}=1$
$\displaystyle \frac{x}{a}+\frac{y}{b}=1$
where, x-intercept $\displaystyle a=-\frac{C}{A}$
y-intercept $\displaystyle b=-\frac{C}{B}$
If $\displaystyle C=0$, then $\displaystyle Ax+By+C=0$ can be written as $\displaystyle Ax+By=0$, which is a line passing through the origin and has zero intercepts on the axes.
Normal Form
The normal form of the equation $\displaystyle Ax+By+C=0$ is,
$\displaystyle x\cos \omega +y\sin \omega =p$
where, $\displaystyle \cos \omega=\pm \frac{A}{\sqrt{A^{2}+B^{2}}}$
$\displaystyle \sin \omega=\pm \frac{B}{\sqrt{A^{2}+B^{2}}}$
$\displaystyle p=\pm \frac{C}{\sqrt{A^{2}+B^{2}}}$
Distance of a Point From a Line
Let $\displaystyle L:Ax+By+C=0$ be a line, whose distance from the point $\displaystyle P(x_{1},y_{1})$ is d.
Draw a perpendicular PM from the point P to the line L.
$\displaystyle d=\frac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}$
Distance between two parallel lines
Let the lines be $\displaystyle L1: y=mx+c_{1}$ and $\displaystyle L2: y=mx+c_{2}$
Line L1 will intersect x-axis at the point $\displaystyle A\left ( -\frac{c_{1}}{m} ,0\right )$
$\displaystyle d=\frac{|c_{1}-c_{2}|}{\sqrt{1+m^{2}}}$
The distance d between two parallel lines $\displaystyle Ax+By+C_{1}=0$ and $\displaystyle Ax+By+C_{2}=0$ is given by,
$\displaystyle d=\frac{|C_{1}-C_{2}|}{\sqrt{A^{2}+B^{2}}}$
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