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How to Parametrize a Line

Two intersecting lines inside a coordinate system

Parametrization is a new way to describe lines and curves in the plane. Normal coordinates are just expressed by numbers for the $x$ - and $y$ -coordinates. When you parametrize a line, you find a parametric equation that expresses the coordinates as functions of new variables like $s$ , $t$ and so on.

One of the main reasons to use parametrization is that it makes it much easier to find out whether two objects are in the same place at the same time. Why is that important? Imagine that you’re going on an airplane. It’s nice to know whether two planes are in the same place at the same time or not, because if they are, it would be catastrophic!

Theory

Parametrization of a Line

Given a point $(x_{0}, y_{0})$ on the line, a vector $\vec{r} = (a, b)$ along the line and a variable $t$ , the parametric equation of the line looks like this:

On vector form:

(x, y) = (x_{0}, y_{0}) + t (a, b)

On coordinate form:

x (t) = x_{0} + a t \land y (t) = y_{0} + b t

x (t) = x_{0} + a t \land y (t) = y_{0} + b t

You can see that the two ways of expressing the parametric equation are closely related. If you set up an equation with all the $x$ -coordinates of the vector form and an equation with all the $y$ -coordinates of the vector form, you will get the parametric equation on coordinate form.

Example 1

Given the parametric equations

$\begin{array}{l} l & : x (t) = 1 + t \land y (t) = 2 - 2 t \\ m & : x (s) = - 3 + s \land y (s) = - 5 + 2 s, \end{array}$

$\begin{array}{l} l & : x (t) = 1 + t \land y (t) = 2 - 2 t \\ m & : x (s) = - 3 + s \land y (s) = - 5 + 2 s, \end{array}$
find the point where the lines $l$ and $m$ intersect.

To find the intersection, create a system of equations where the $x$ -coordinate of $l$ is equal to the $x$ -coordinate of $m$ , and the $y$ -coordinate of $l$ is equal to the $y$ -coordinate of $m$ :

\begin{array}{l} 1 + t & = - 3 + s \\ t & = - 4 + s \\ 2 - 2 t & = - 5 + 2 s \\ ⇓ \\ - 2 (- 4 + s) & = - 5 + 2 s \\ 8 - 2 s & = - 5 + 2 s \\ 13 & = 4 s \\ s & = \frac{13}{4} \\ t & = - 4 + s \\ ⇓ \\ t & = - 4 + \frac{13}{4} \\ t & = - \frac{3}{4} \end{array}

\begin{array}{l} 1 + t & = - 3 + s & 2 - 2 t & = - 5 + 2 s \\ t & = - 4 + s & - 2 (- 4 + s) & = - 5 + 2 s \\ 8 - 2 s & = - 5 + 2 s \\ 13 & = 4 s \\ s & = \frac{13}{4} \\ t & = - 4 + \frac{13}{4} \\ t & = - \frac{3}{4} \end{array}

You can find the intersection by inserting the values you found for

s

t

back into one of the parametric equations:

\begin{array}{l} x (\frac{- 3}{4}) & = 1 + \frac{- 3}{4} = \frac{1}{4} \\ y (\frac{- 3}{4}) & = 2 - 2 \cdot \frac{- 3}{4} = \frac{7}{2} \end{array}

That means the lines $l$ and $m$ intersect in the point $(\frac{1}{4}, \frac{7}{2})$ . Note! You can also insert for $s$ in the other parametric equation, which will give you the same point. If you have the time, it’s always smart to do that just to check whether you have made any mistakes or not!

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