For two position vectors $P_1$ and $P_2$ ; a parametric representation between them is:

$P(t)=P_1+(P_2-P_1)t$

where

$0\le t\le 1$-----where t is the parameter

$P(t)$-----point, as a function of t

Proof:-

In polynomial form $\because$ we have a linear equality,

$P(t)=a_0+a_1t$..........where $a_0$ & $a_1$ are the unknown constants

B.C.'s (Boundary Conditions)

At $t=0; P=P_1$

and at $t=1;P=P_2$

$\therefore P_1=a_0$,

$P_2=P_1+a_1 \Rightarrow a_1=(P_2-P_1)$.

$\therefore P(t)=P_1+(P_2-P_1)t$

Now, $P(t)=[x(t) \ \ \ y(t)]$

$\therefore x(t)=x_1+(x_2-x_1)t$

$y(t)=y_1+(y_2-y_1)t$

where $0\le t \le 1$

E.g. For the position vectors $P_1[1 \ \ \ 2]$ and $P_2[4 \ \ \ 3];$ determine the parametric representation of the line segment between them. Also determine the slope and tangent vector of the line segment.

Solution:- $P(t)=P_1(t)+(P_2-P_1)t$ $\forall$ $0\le t\le 1$

$=[1 \ 2]+([4 \ 3]-[1 \ 2])t$

$=[1 \ 2]+[3 \ 1]t$ $\forall$ $0\le t\le 1$

For x & y components; the parametric representation are:-

$x(t)=1+3t$

& $y(t)=2+t$

where $0\le t\le 1$

The target vector is obtained by differentiating P(t)

$\therefore P'(t)=[x'(t) \ \ \ y'(t)=[3 \ \ \ 1]$

or $V_t=3i+j$

The slope of the line segment is:-

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y'(t)}{x'(t)}=\frac{1}{3}$