# How do you convert (y = 2 + x) to parametric equation?

Jul 29, 2016

$l \to \left\{\begin{matrix}x = 0 + \lambda \\ y = 2 + \lambda\end{matrix}\right.$ for $\lambda \in \mathbb{R}$

#### Explanation:

The general line equation in ${\mathbb{R}}^{2}$ is

$l \to a x + b y + c = 0$

this equation can be arranged as

$l \to \left\langle\vec{n} , p - {p}_{0}\right\rangle = 0$ where

$\vec{n} = \left\{a , b\right\}$, $p = \left\{x , y\right\}$ a line generic point, and ${p}_{0} = \left\{{x}_{0} , {y}_{0}\right\}$ a line given point.

The determination of ${p}_{0}$ is straightforward,
comparing

$a x + b y + c = 0$ with
$a \left(x - {x}_{0}\right) + b \left(y - {y}_{0}\right) = 0$ concluding

$- a {x}_{0} - b {y}_{0} = c$

then supposing that $b \ne 0$ and fixing ${x}_{0} = 0$ we get ${y}_{0} = - \frac{c}{b}$ so

${p}_{0} = \left\{0 , - \frac{c}{b}\right\}$

Now, the parametric representation.

We have a line point which is ${p}_{0}$ and an orthogonal vector $\vec{n}$ to the line direction. We need a vector $\vec{v}$ in the line direction, so choosing $\vec{v}$ such that $\left\lVert \vec{v} \right\rVert > 0$ and $\left\langle\vec{n} , \vec{v}\right\rangle = 0$. Choosing $\vec{v} = \left\{b , - a\right\}$ then $\left\langle\vec{n} , \vec{v}\right\rangle = b a - a b = 0$.

Finally, the parametric line equation is

$l \to p = {p}_{0} + \lambda \vec{v}$ with $\lambda \in \mathbb{R}$

In the present case

$l \to - x + y - 2 = 0$

${p}_{0} = \left\{0 , - \frac{c}{b}\right\} = \left\{0 , - \left(\frac{- 2}{1}\right)\right\}$
$\vec{n} = \left\{- 1 , 1\right\} \to \vec{v} = \left\{1 , 1\right\}$

then

$l \to \left\{\begin{matrix}x = 0 + \lambda \\ y = 2 + \lambda\end{matrix}\right.$

Jul 29, 2016

see below

#### Explanation:

parametrising a straight line is a bit of a waste of time unless you have an idea as to why you would want to do that.

or, put another way, the idea behind parameterisation is that you reduce the independent variables. in Cartesian x-y, it really kinda misses the point.