# How do you find the parametric equations for a line segment?

##### 1 Answer
Sep 6, 2014

The line segments between $\left({x}_{0} , {y}_{0}\right)$ and $\left({x}_{1} , {y}_{1}\right)$ can be expressed as:
$x \left(t\right) = \left(1 - t\right) {x}_{0} + t {x}_{1}$
$y \left(t\right) = \left(1 - t\right) {y}_{0} + t {y}_{1}$,
where $0 \le q t \le q 1$.

The direction vector from $\left({x}_{0} , {y}_{0}\right)$ to $\left({x}_{1} , {y}_{1}\right)$ is
$\vec{v} = \left({x}_{1} , {y}_{1}\right) - \left({x}_{0} , {y}_{0}\right) = \left({x}_{1} - {x}_{0} , {y}_{1} - {y}_{0}\right)$.
We can find any point $\left(x , y\right)$ on the line segment by adding a scalar multiple of $\vec{v}$ to the point $\left({x}_{0} , {y}_{0}\right)$. So, we have
$\left(x , y\right) = \left({x}_{0} , {y}_{0}\right) + t \left({x}_{1} - {x}_{0} , {y}_{1} - {y}_{0}\right)$,
which simplifies to:
$\left(x , y\right) = \left(\begin{matrix}1 - t {x}_{0} + t {x}_{1} \\ 1 - t {y}_{0} + t {y}_{1}\end{matrix}\right)$,
where $0 \le q t \le q 1$.