# Parametric forms for lines and vectors

*parametric form*for a line occurs when we consider a particle moving along it in a way that depends on a parameter \(\normalsize{t}\), which might be thought of as time. Thus both \(\normalsize{x}\) and \(\normalsize{y}\) become functions of \(\normalsize{t}\). The simplest parameterisation are linear ones. In this step we

- look at parameterisations of the simplest curves: lines
- show how to go from parametric equations to Cartesian equations
- introduce vectors.

## Parameterisations of a line

A simple example of a parameterisation is a linear parameterisation: \[\Large{p(t)=[t+1, 2t-3]}.\]As \(\normalsize{t}\) varies, the point \(\normalsize{p(t)}\) moves along a line. When \(\normalsize{t=0}\), \(\normalsize{p(0)=[1,-3]}\), when \(\normalsize{t=1}\), \(\normalsize{p(1)=[2,-1]}\), and when \(\normalsize{t=2}\), \(\normalsize{p(2)=[3,1]}\). Clearly also fractional values of \(\normalsize{t}\) are allowed.How do we go from a parametric form for a line to one of the more usual Cartesian forms involving \(\normalsize{x}\) and \(\normalsize{y}\)? We eliminate the parameter \(\normalsize{t}\). For example for the parametric equation \(\normalsize{p(t)=[t+1, 2t-3]}\) we have \[\Large{x=t+1}\] \[\Large{y=2t-3}\]so that \(\normalsize{y=2(x-1)-3}\) or \(\normalsize{y=2x-5}\). Please check that this is the same line – with a different equation.Q1(E): Find a Cartesian equation for the parametric line \({\normalsize p(s) = [2s+1, 4s-3]}\).

## How to go from a Cartesian equation to a parametric form?

Q2(E): Find another parameterisation of \(\normalsize y=3x+2\) by setting \(\normalsize x=3t\).

## Using vectors to parameterise lines

**vector**is a directed line segment, or geometrically a difference between two points in the plane. We will denote vectors by round brackets, such as \(\normalsize{v=(1,4)}\). This represents a relative displacement of \(\normalsize{1}\) to the right in the \(\normalsize x\)-direction and \(\normalsize{4}\) up in the \(\normalsize y\)-direction. We agree that vectors are not fixed in place, as they are only determined in a relative sense.

## Answers

A1.This is also \(\normalsize{y=2x-5}\).A2.Another parameterisation is \(p(t) = [3t, 9t+2]\).

#### Maths for Humans: Linear, Quadratic & Inverse Relations

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