The shortest distance between any two distinct points is the line segment joining them.How can I see why this is true?

Question

On a euclidean plane, the shortest distance between any two distinct points is the line segment joining them. How can I see why this is true?

Answer 1

Every now and then it's nice to nuke a mosquito.

Let's assume that the path connecting two points $(a,y(a))$ and $(b,y(b))$ can be expressed as a function, and the curve $C(x)$ is given by $C(x) = (x,y(x))$. Then we will proceed using the Calculus of Variations.

The derivative of $C$ wrt $x$ is $(1, y')$, and the functional we want to minimize is the length of the curve $L = \int \|C'\|dx = \int_a^b\sqrt{1 + y' ^2} dx$. If we take $f(x,y,y') = \sqrt{1 + y'^2}$, we get that $\frac{df}{dy} = 0, \frac{df}{dy'} = \frac{y'}{\sqrt{1 + y'^2}}$. Then the Euler-Lagrange equation, sometimes referred to as the fundamental equation of the Calculus of Variations, says exactly that $\dfrac{d}{dx} \left( \frac{y'}{\sqrt{1 + y'^2}}\right) = 0$, which is exactly that $y'$ is a constant.

Thus, if the path connecting the two points is expressible as a function, then the shortest such path is given by a straight line.

EDIT I was certain that someone was in the middle of writing an answer when I typed my tongue-in-cheek response (as so often happens), but as I now see that there is more to add, allow me to extend my answer

The problem here is that we must first define "distance." In the standard Euclidean Plane, the distance between two points is defined to be the length of the line segment between them. So we can drop the word 'shortest' and say that "The distance between any two distant points is the length of the line segment joining them."

Presumably, you want to know that going along any other path will be at least as long. One way of 'seeing this' is that you can approximate any curve with a polygonal path, and these satisfy the triangle inequality, which will make the path longer.

Answer 2

Let $\gamma(s)$ be a continuous curve in the plane with end-points $\gamma(0) = a$ and $\gamma(1) = b$. Using the Euler-Lagrange equations, the only stationary solution is $\gamma(s) = bs + (1 - s)a$, which is a line connecting the two end-points. See also this.