Solution 1:

Like this:

$\hskip1.7in$ enter image description here

Solution 2:

Five lines ten points

This is an alternative (sorry diagram is clunky)

Solution 3:

5 lines times 4 soldiers on a line equals 20 = two times 10 soldiers available. This suggests that every soldier belongs to two lines.

Draw $n$ lines on the plane such that no two are parallel, and no three intersect in one point.

You can always do that: if you already have $n-1$ lines then there are finite number of slopes of those lines and finite number of points of intersection -- choose a new slope not equal to any previous and draw the line with this slope not going through any previous points of intersection.

  1. Each line contains exactly $n-1$ points of intersection with other lines.

  2. There are $\frac{n(n-1)}{2}$ intersection points in total.

Now, if you put a soldier at every point of intersection, then there are $\frac{n(n-1)}{2}$ soldiers arranged in $n$ lines, each containing $n-1$ soldier. For $n=5$ you get the answer: any such configuration of 5 lines would work.

Solution 4:

A more irregular looking solution.

enter image description here