Finding the coordinates of R such that PR+RQ is minimum Let P$=(1,1)$ and Q$=(3,2)$. Find the point R on the $x$-axis such that PR$+$RQ is minimum.

Let the coordinates of R be $(h,k)$.
For PR+RQ to be minimum, PRQ would have to be a straight line. But R lies on the $x$-axis,
So $k=0$ and $h=\frac{1+3}{2}=2$. So R(2,0)
But the answer happens to be ($\frac{5}{3},0)$
 A: 

By Fermat's principle of minimum time of path $PR+RQ $ or total path length ( hope you can set it up else we would help) as a function of point $R$ abscissa for independent variable  it can be shown by differentiation that the line joining image P' and Q (or image of Q and  P) intersects x-axis at the required bounce off point $R$. On that basis,
$$ \dfrac{y-2}{x-3} = \dfrac{0-2}{x-3} = \dfrac{3}{2}, x_R= \frac{5}{3},y_R=0 ;\; $$
A Geogebra dimensioned graph/ sketch has been updated to the earlier rough quick one,please note note that the x-axis is a mirror, $P'RQ$ is straight, $PR+RQ= 2a$, the major axis.
BTW, $P,Q$ are foci of an ellipse and $R$ is tangent point to x-axis.
If it is required to reflect at y-axis, imagine a slightly expanded ellipse on the same foci touching y-axis.
A: Notice that the $x$-axis is not between the two points -- both of them are in the upper plane. So you are looking for the intersection of the $x$-axis and the line between $P$ and $Q'$, where $Q'$ is the reflection w.r.t. the $x$-axis.
You can have a look at pictures given here to get the idea why this works: Minimizing the length of wire between two poles?
If a point $(h,k)$ is on the line given by the points $P=(1,1)$ and $Q'=(3,-2)$, then you have
\begin{align*}
h&=1+2t\\
k&=1-3t
\end{align*}
If you set $k=0$, you should be able to calculate $t$ and $h$.
