It is a well-known face, first enunciated by Archimedes, that the shortest distance between two points in a plane is a straight line. However, suppose that we wish to demonstrate this result from first principles. Let us consider the length l, of various curves y(x) which run between two fixed points A and B in a plane, as illustrated in Figure below.

$$\therefore l = \int\limits_{A}^{B} \sqrt{dx^2 + dy^2} = \int\limits_{A}^{B} \sqrt{1 + y'^2}dx.....(1) \\where \hspace{0.25cm} y' = \frac{dy}{dx}$$

Now, in order to find the shortest path between points A and B we need to minimize the functional l with respect to small variations in the function y(x) subject to the constraint that the end points A and B remain fixed.

$$\therefore F = \sqrt{1 + y'^2}.........(2)$$

Since F does not contain x explicitly, we have

$$F - y' \frac{\partial F}{\partial y'} = c.........(3) \\ \therefore \frac{\partial F}{\partial y'} = \frac{y'}{\sqrt{1 + y'^2}}........(4)$$

Substituting eqn(2) and eqn(4) in eqn(3) we get,

$$\sqrt{1 + y'^2} - y' * \frac{y'}{\sqrt{1 + y'^2}} = c \\ \frac{1 + y'^2 - y'^2}{\sqrt{1 + y\^2}} = c \\ \therefore \frac{1}{\sqrt{1 + y'^2}} = c \\ \therefore \frac{1}{1 + y'^2} = c^2$$

$$\therefore y' = \frac{\sqrt{1 - c^2}}{c} \\ \because y ' = constant$$

$\therefore$ It is the equation of a straight - line. Thus, the shortest distance between two fixed points in a plane is indeed a straight line.