Different Forms of an Equation of a Straight Line

Given any two independent pieces of information about a line from among points, slope, and intercepts, its equation can be written directly. There are four standard combinations, plus two further special forms:

1. Slope and intercept form. A line through the origin with slope $m$ is $y=mx$. In general, a line with slope $m$ and $y$-intercept $b$ ($b\ne0$) is

   $y=mx+b.$

   Special cases: if $b=0,\ m\ne0$, the line is $y=mx$ (through the origin); if $b=0,\ m=0$, the line is the $x$-axis, $y=0$; if $b\ne0,\ m=0$, the line is horizontal, $y=b$ (parallel to the $x$-axis).
2. Point-slope form. For a line of slope $m$ through a known point $A(x_1,y_1)$, any other point $P(x,y)$ on the line satisfies $m=\dfrac{y-y_1}{x-x_1}$, i.e.

   $y-y_1=m(x-x_1).$

   Because the slope of a vertical line is undefined, this form cannot express a line through $(x_1,y_1)$ parallel to the $y$-axis — but that line is simply $x=x_1$ (every point on it shares the same $x$-coordinate), so no real difficulty arises.
3. Two-point form. For distinct points $(x_1,y_1),(x_2,y_2)$ with $x_2\ne x_1,\ y_1\ne y_2$, the slope is $m=\dfrac{y_2-y_1}{x_2-x_1}$; substituting into the point-slope form and rearranging gives

   $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1},$

   equivalently the determinant form $\begin{vmatrix}x-x_1 & y-y_1\\ x_2-x_1 & y_2-y_1\end{vmatrix}=0$.
4. Intercept form. If the $x$-intercept is $OA=a$ and the $y$-intercept is $OB=b$ (both nonzero), the line passes through $A(a,0)$ and $B(0,b)$; applying the two-point form and simplifying gives

   $\frac{x}{a}+\frac{y}{b}=1.$

   A line through the origin, or a horizontal or vertical line, violates the 'both intercepts nonzero' requirement and so cannot be written in this form — but this form is often the quickest for sketching a line's graph, since both axis crossings are read off immediately.
5. Normal form. Let $p$ be the length of the perpendicular $OP$ dropped from the origin to the line, making angle $\alpha$ with the $x$-axis, and let the line meet the axes at $A,B$. In right triangles $OPA,OPB$: $\cos\alpha=OP/OA=p/OA$ and $\sin\alpha=\cos(\pi/2-\alpha)=OP/OB=p/OB$, so $OA=p/\cos\alpha$, $OB=p/\sin\alpha$. Substituting these as the intercepts $a,b$ in the intercept form and simplifying gives

   $x\cos\alpha+y\sin\alpha=p,$

   valid for every position of the line provided $p$ is always taken positive and $\alpha$ is always measured from the positive $x$-axis.
6. Parametric (symmetric) form. For a line through a fixed point $Q(x_1,y_1)$ making angle $\theta$ with the $x$-axis, let $P(x,y)$ be any point on the line at signed distance $r$ from $Q$ (positive on one side of $Q$, negative on the other). Dropping perpendiculars from $Q,P$ to the $x$-axis and constructing the right triangle between them gives $x-x_1=r\cos\theta$ and $y-y_1=r\sin\theta$, i.e.

   $\frac{x-x_1}{\cos\theta}=\frac{y-y_1}{\sin\theta}=r \qquad (\text{equivalently } x=x_1+r\cos\theta,\ y=y_1+r\sin\theta).$

   Here $r$ is called the parameter; every point on the line corresponds to a unique value of $r$, positive on one side of $Q$ and negative on the other, so this form is especially convenient for 'a point on the line at a given distance from a known point' problems, since $r$ can simply be set equal to that given distance (with either sign). Summary table (with the general equation added as a seventh, all-encompassing form):