This is a cross-post of a question on MathOverflow where it didn't get much attention: https://mathoverflow.net/questions/337892/how-to-find-a-point-on-a-line-that-minimizes-sum-of-distances-from-three-given-p
Let there be given three points $x_1$,$x_2$,$x_3$ and a line $l$ on a plane. Does there exist an explicit method of finding such a point $p$ on $l$, that sum of distances of $p$ from $x_1$,$x_2$,$x_3$ is possibly minimal? Such result for two given points and a line is trivial and known as the Heron's theorem commonly used in optics to find a path of a ray of light.
All I know about the solution is that if $p$ minimizes the sum of distances, then
$$\cos{\alpha}+\cos{\beta}+\cos{\gamma}=0$$
The equation above comes directly from the derivative of the function of sum of distances if we assume that the line has equation $y = 0$:
$$\frac{d}{dx} \sum_{i=1}^3 \sqrt{(x-x_i)^2+(0 – y_i)^2} = 0$$
$$\sum_{i=1}^3 \frac{x-x_i}{\sqrt{(x-x_i)^2+(y_i)^2}} = 0$$
$$\cos{\alpha}+\cos{\beta}+\cos{\gamma}=0$$
However, that is not really helpful neither for construction nor finding the solution analytically. Some rare situations may be solved using Brianchon's theorem.
That is, if we add symmetrical reflections of $x_1$,$x_2$,$x_3$ on the opposite sides of $l$ and the six points are vertices of a hexagon circumscribing a conic section, then the intersection of the main diagonals minimizes the sum of distances. However it is far from general solution.
Best Answer
As JJacquelin answered, Newton method is the simplest way to solve your problem.
As usual, you need a guess. You can get it minimizing first the sum of the squared distances that is to say $$\frac{d}{dx} \sum_{i=1}^3 \big({(x-x_i)^2+(0 - y_i)^2}\big) = 0\implies x=\frac 13\sum_{i=1}^3 x_i$$ For the test case, this gives $x_0=6.83$.
Now, for the real problem, Newton method will generate the following iterates $$\left( \begin{array}{cc} n & x_{(n)} \\ 0 & 6.8333333 \\ 1 & 6.3992290 \\ 2 & 6.4017089 \\ 3 & 6.4017090 \end{array} \right)$$
Edit
Making the problem more general for $n$ points, the iterates of Newton method are given by
$$x_{(n+1)}=x_{(n)}-\frac{\sum _{i=1}^n \frac{x_{(n)}-x_i}{\left(x_{(n)}^2-2 x_{(n)} x_i+x_i^2+y_i^2\right)^{1/2}}}{\sum _{i=1}^n \frac{y_i^2}{\left(x_{(n)}^2-2 x_{(n)} x_i+x_i^2+y_i^2\right)^{3/2}}}\qquad \text{with} \qquad x_{(0)}=\frac 1n\sum_{i=1}^n x_i$$ which should converge in a couple of iterations.