Horizontal and Vertical Lines
If a vertical line L is at a distance b from the y-axis then equation of L is either x=a or x=-a.
Point-slope Form
Suppose $\displaystyle P_{0}(x_{0},y_{0})$ is a fixed point on a non-vertical line L, whose slope is m.
Let $\displaystyle P(x,y)$ be an arbitrary point on L.
$\displaystyle m=\frac{y-y_{0}}{x-x_{0}}$
$\displaystyle y-y_{0}=m(x-x_{0})$
Thus, the point (x,y) lies on the line with slope m through the fixed point $\displaystyle (x_{0},y_{0})$, if and only if, its coordinates satisfy the equation $\displaystyle y-y_{0}=m(x-x_{0})$
Two-point Form
Let the line L passes through two given points $\displaystyle P_{1}(x_{1},y_{1})$ and $\displaystyle P_{2}(x_{2},y_{2})$.
Let $\displaystyle P(x,y)$ be a general point on L.
$\displaystyle y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$
Slope-intercept Form
Case I: Suppose a line L with slope m cuts the y-axis at a distance c from the origin.
The equation of L is,
$\displaystyle y-c=m(x-0)$
$\displaystyle y=mx+c$
Case II: Suppose line L with slope m makes x-intercept d.
Then equation of L is,
$\displaystyle y=m(x-d)$
Intercept Form
Suppose a line L makes x-intercept a and y-intercept b on the axes.
L meets x-axis at the point (a,0) and y-axis at the point (0,b).
$\displaystyle \frac{x}{a}+\frac{y}{b}=1$
Normal Form
Let L be the line, whose perpendicular distance from origin O be OA=p and the angle between the positive x-axis and OA be $\displaystyle \angle XOA=\omega$.
The possible positions of line L in the Cartesian plane are,
$\displaystyle MA=p\sin \omega$
Line L is perpendicular to OA.
The slope of line = $\displaystyle -\frac{1}{slope of OA}$
$\displaystyle =-\frac{1}{\tan \omega }=-\frac{\cos \omega }{\sin \omega }$
The equation of the line is,
$\displaystyle x \cos \omega+y\sin \omega=p$
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