Parameterizing the equation of a line in 2-D

Question

I'm attempting to read Emil Artin's little book on the Gamma function with a borderline adequate background at best.

In order to make it past the first page I need to understand what a convex function is. That leads me to the inequality $tf(x_1)+(1−t)f(x_2)≥ f(tx_1+(1−t)x_2)$. But in order to understand that I need to understand the parameterization of a line between two points on a function, namely $r(t) = tx_1 + (1-t)x_2$. In short, I can't fully intuit how this is the parametric equation of a line.

I'm modestly familiar with parameterized lines in 3-D. So if you were to ask me to parameterize a line in 2-D I'd assume I'd need a beginning point and a directional vector.

Let's take the points $(1,1)$ and $(4,2)$. A vector between from $(1,1)$ and $(4,2)$ would be $<3,1>$. So a parametric line would be something like $r(t) = <1,1> + t<3,1> = (1 + 3t, 1 + t)$, no? But what does that have to do with parameterization of the form $r(t) = tx_1 + (1-t)x_2$?

In a nutshell:

1. Sanity check: When describing the parametric line between points on a convex function, $x_1$ and $x_2$ are the $x$-coordinates of two separate points, correct? (As opposed to, say, $x_1$ being the $x$-coordinate and $x_2$ being the $y$-coordinate of a single given point).

2. Examples from scratch are tremendously helpful. What is the parametric equation of the line segment from $(1,1)$ to $(4,2)$ using the sort of parameterization used in describing convexity? Please be descriptive and show how a couple of additional points on the line are computed for good measure if possible.

3. Why $t$ and $1 -t$, as opposed to, say, $t$ and $1 + t$? How are these parameters derived from first principles?

I'm aware that there are many questions on parametric lines already answered. But after sifting through many for the past couple of hours I did not find one that fully explicated what I'm after.

Thank you.

Answer 1

Of course moments after saying I can't find an answer I stumble upon the first truly helpful post. So borrowing from it, I believe I can answer my own questions.

1. No, $x_1$ and $x_2$ are points on the x-y coordinate system, and in the interest of reduced ambiguity it would help to write them, rather, as $u_1 = (x_1, y_1)$ and $u_2 = (x_2, y_2)$.

2. $r(t) = tu_1 + (1 - t)u_2$ is equivalent to $r(t) = (tx_1+(1−t)x_2, ty_1+(1−t)y_2)$. So we would naturally parameterize the line through $(1,1)$ and $(4,2)$ as $r(t) = (t + (1-t)4, t + (1-t)2) = t(1,1) + (1-t)(4,2)$. So, for example, when $t= 0.5$ we have $(0.5 + 0.5(4), 0.5 + 0.5(2)) = (2.5, 1.5)$, another point on the line.

3. The coefficients $t$ and $1-t$ are chosen so that the sum is $1$, which guarantees that points stay on the line.

The final point is the only place where I'm left wanting a tad more. This makes sense, but I wouldn't mind seeing a quick proof that coefficients summing to $1$ means we stay on the same line.

Answer 2

$x=1t+4(1-t)=4-3t,y=1t+2(1-t)=2-t$. As $t$ goes from 0 to 1, the point goes from (4,2) to (1,1).

Answer 3

In order to match your parametrization one neeeds to start from $t(4,2)+(1-t)(1,1).$ Then adding these gives $(4t-t+1,2t-t+1)=(1+3t,1+t)$ as you obtained using the direction vector method.