Where should a water pump be placed to minimize the length of the pipe?

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The problem is as follows:

Two buddies Reggie and Mathias have their homes nearby a fresh water river as indicated in the picture from below. Assuming they have only one available water pump which they can use. Where must be located this pump so that the pipes length is the minimum to provide water to both homes?.

Sketch of the problem

The choices given in my workbook are as follows:

$\begin{array}{ll} 1.&\textrm{at 25 m from A}\\ 2.&\textrm{at 25.71 m from A}\\ 3.&\textrm{at 30 m from B}\\ 4.&\textrm{at 26 m from B}\\ \end{array}$

I'm totally lost in this geometry problem. Can someone help me here?. I believe that the catch here is to make a right triangle or some sort of isosceles but I'm not sure of the rationale or most importantly the reason why?. Therefore I'm asking for assistance.

I would really like to provide more effort into this, but I'm stuck. I don't know what else can I do?.

Can you please include some drawing in your answer as a justification. I really beg someone include a drawing becuase I can't spot where should this pump be located.

Please help because I can't figure this out.

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Imagine that line segment AB is a mirror and that Mathias shines a light from his house to Reggie's house by reflecting it off the mirror. We know from physics that the light ray will take the shortest path, and that the angle of incidence is equal to the angle of reflection. That is, if Reggie is at point $R$ and Matthias is at point $M$, then in order for the light to go from $M$ to $R$, it must strike the mirror at a point $X$ such that $\angle MXA = \angle RXB$. But then triangles $MXA$ and $RXB$ are similar, so $X$ must divide segment $AB$ in the ratio $3:4$. That is, $X$ is $\frac{180}7$ meters from $A$ and the correct answer is $2$.

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Let AB be the perpendicular bisector of MM' and RR'. MR',M'R and AB meet at P. Here is where you should build the water pump.

For any other point along the bank, for example X, MX+XR =M'X+XR>M'R=MP+PR.

Set $\angle R=\angle M=\theta$,

30 $\tan\theta+40\tan \theta=60$, $\tan \theta=\frac{6}{7}$,

$\theta = 40.6^\circ$, MP= $30\tan 40.6^\circ$=25.71m$