Equations of lines

There are many ways of representing a mathematical line. You are probably most familiar with point-y-intercept form,

y=mx+b\displaystyle y = m x + b.

That equation is useful when you know the slope and the y-intercept. Also, that form is easy to express in function notation, like f(x)=mx+b\displaystyle f{\left ( x \right )} = m x + b, and it fits nicely with the other families of functions. Another way of describing a line is with standard form:

ax+by+c=0\displaystyle a x + b y + c = 0.

This form gives you both the intercepts, but most of the time it is inconvenient. Yet another type of line equation is point-slope form:

yy0=m(xx0)\displaystyle y - y_{0} = m \left ( x - x_{0} \right ).

Not surprisingly, this is useful when you know a point and the slope.

Those three equations have something in common: they relate x to y. If you solve for y (already done for you in point-y-intercept form), you can choose an x-value and the equation will tell you the y-value. This isn’t the only way of doing it, as we will see. In particular, it doesn’t scale well to three-space; our new method will.

We are going to explore another triplet of line equations, starting with the vector equation of the line. To describe a line with vectors, we need an initial point that falls on the line—we’ll call it point A. From point A, we need a vector that tells us which way the line goes—we’ll call this m\displaystyle \vec{m}. Here’s what this might look like:

xyA(1,−2)m=[3,2]P110−1−222345678
A line defined by an initial point and a direction vector

To get from point A to some point P elsewhere on the line, all we have to do is add a scalar multiple of the direction vector:

OP=OA+tm\displaystyle \overrightharpoon{O P} = \overrightharpoon{O A} + t \vec{m}.

The value of t controls how far along the line we go from A. If it is negative, we go in the opposite direction. Instead of using origin-to-point notation, we usually use r\displaystyle \vec{r}, which is implied to be a position vector (the r stands for radius, and a position vector is a radius from the origin). Here is the conventional way of writing the vector equation of the line:

r=r0+tm,tR\displaystyle \vec{r} = \vec{r}_{0} + t \vec{m} , \qquad t \in \mathbb{R}.

Suppose we substitute for vectors in R2\displaystyle \mathbb{R}^{2} with variables as components:

[x,y]=[x0,y0]+t[a,b],tR\displaystyle \left [ x , y \right ] = \left [ x_{0} , y_{0} \right ] + t \left [ a , b \right ] , \qquad t \in \mathbb{R}.

Breaking this up into two equations gives us the parametric equations of a line (two equations because there are two components):

x=x0+taandy=y0+tb,tR\displaystyle x = x_{0} + t a \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad y = y_{0} + t b , \qquad t \in \mathbb{R}.

There is an important difference between this form and the earlier ones. The other equations relate x and y such that we can choose one an obtain the other. With the parametric equations, we choose a value for the free parameter t. We plug this into the equations and they tell us x and y. Instead of x being related to y, they are both independently controlled by t.

If we solve the parametric equations for t, we get the symmetric form:

t=xx0a=yy0b\displaystyle t = \frac{x - x_{0}}{a} = \frac{y - y_{0}}{b}.

This form makes it easy to test if a particular point falls on the line: just evaluate both sides and see if they produce the same t-value.