How to find the parametric representation of a line?

Question

I found these equations online given two points.

x = (1 - t)x1 + tx2

y = (1 - t)y1 + ty2

Where do these equations come from?

And how can you find the parametric representation of a line given y=ax+b?

Answer 1

They represent a weighted average of the point $P = (x_1, y_1)$ and the point $Q =(x_2, y_2)$, with weights $t$ and $1-t$, which have the property that they sum to 1, which happens to be exactly what's needed to make the weighted average be on the line between the two points.

One way to see this is that if you moved each of the points $P$ and $Q$ to the right one unit (i.e., increased $x_1$ and $x_2$ both by 1), then the $x$-coordinate for your equation would increase by $t$ and by $1-t$, hence would increase by a total of 1. The same thing goes for $y$. We say that the construction is "translation invariant" (or better "translation equivariant").

So if the formulas give the right answer (i.e. a line) for $P$ and $Q$, they also give the right answer for the same two points, displaced by some amount.

Apply that idea to displace both points by $-x_1$ in $x$ and then by $-y_1$ in $y$. Your new points are then $$ P' = (0, 0)\\ Q' = (x_2 - x_1, y_2 - y_1). $$ Let's call the coordinates of $Q'$ $c$ and $d$ instead, so $$ c = x_2 - x_1 \\ d = y_2 - y_1 $$ What does the formula give for points on the line between $P'$ and $Q'$? Answer: $$ x(t) = (1-t) 0 + tc = tc \\ y(t) = (1-t) 0 + td = td $$ Since these point satisfy the line equation $$ y = cx $$ which is satisfied by both $P'$ and $Q'$, the formula is clearly generating a line between $P'$ and $Q'$...which means it must also have generated a line between $P$ and $Q$.

As for converting: Pick $(x_1, y_1) = (0, b)$ and $(x_2, y_2) = (1, a+b)$; that'll work just fine.