Creating a parametric Equation when given the points of a collinear line?

Question

$(-70, 3)$, $(88, 81)$, and $(246, 159)$ are three collinear points. Write parametric equations for $x$ and $y$. (In other words, write equations that produce points when $t$-values are assigned.)

And it is regarding Parametric Equations and Geometry.

This was from an old test that I never got to talk with my teacher about and I think it will be on our next test in some variation.

Answer 1

Take two of those points, $A$ and $B$, and set $f(t)=A+t(B-A)$. Then split into equations for x and y.

Answer 2

Let $P=(-70,3)$ and $Q=(88,81)$, and let $\vec{a}=\vec{PQ}=\langle158,78\rangle$.

Using the coordinates of $P$, we obtain $x=-70+158t, y=3+78t$ as parametric equations for the line.

Answer 3

Let two points $(x_1,y_1)$ and $(x_2,y_2)$ be given.

Parametric equation of a straight line:

$1) x = a + bt$, and

$2) y= c + dt$, where

$t$ is the parameter, $a,b,c$ and $d$ are constants that are determined by the two points.

First point, $(x_1,y_1)$:

$ x_1 = a + bt$, and $ y_1 = c+ dt$ .

Choose, $t=0$ at this point to get:

$a = x_1$, and $c = y_1$ .

Second point, $(x_2,y_2)$:

$x_2 = x_1 +bt$, and $y_2 = y_1 + dt$;

Choose $t = 1$ to get:

$b = x_2 - x_1$, and $d = y_2 - y_1$.

Finally the parametric equations 1) and 2):

1) $x = x_1 + (x_2 - x_1)t$ , and

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

Check:

$t = 0: x = x_1, y = y_1$, and

$t = 1: x = x_2, y = y_2$.

The straight line passes through the given points.

Note: This parametric representation is not unique.

Any other choice of parameter of the form:

$t' = \alpha t + \beta \,$, $ \, \alpha ,\beta$ constants, will do.

This gives us the freedom to choose $t = 0$ at $(x_1,y_1)$, and $t= 1$ at $(x_2,y_2)$.