Parametrizing a Line Segment

A straight-line trip has no turning and no speeding up, so two numbers pin it down completely: where you start, and how fast x and y each change per unit time. Everything else is just adding distance traveled to the starting position. That is why a segment from (x0, y0) to (x1, y1) only needs one starting point and two constant rates, one for x and one for y.

To make the trip take exactly one second, build the rates from the total displacement. Set x(t) = x0 + t(x1 - x0) and y(t) = y0 + t(y1 - y0). At t = 0 you are sitting at (x0, y0). At t = 1, the t-terms have grown to their full size, landing you exactly at (x1, y1). Any t between 0 and 1 gives a point that fraction of the way along the segment, which is exactly what linear interpolation means.

This connects directly to slope. The rate of change in y is (y1 - y0) and the rate of change in x is (x1 - x0); the particle's path is just the line through the two points traced at constant speed. Divide those two rates and you get (y1 - y0)/(x1 - x0), the ordinary slope of the line. So a parametrized segment is really slope and a starting point in disguise, split into separate clocks for x and y, which is the same idea that later lets you parametrize curves that are not lines by letting those rates vary with t.