Determine the length of the shortest path from A to B consisting of the line segments AP, P Q? Let $A = (0, 1)$ and $B = (1, 1)$ in the plane $\mathbb{R}^2$ . Determine the length of
the shortest path from $A$ to $B$ consisting of the line segments $AP$, $PQ$ and
$QB$, where $P$ varies on the $x$-axis between the points $(0, 0)$ and $(1, 0)$ and $Q$
varies on the line $y = 3$ between the points $(0, 3)$ and $(1, 3)$.
I was trying this question many times, but I could not able to solve it. I was  using the distance formula, but I didn't  get correct value..
If anybody help me I would be very thankful to him..
 A: Let $M(0,-1)$ and $N(1,5)$. 
Hence, the length of $MN$ is an answer. 
Let $O(0,0)$, $C(0,3)$, $D(1,3)$, $MN\cap CD=\{Q'\}$ and $MN\cap x-axis=\{P'\}.$ 
Thus, since $DB=DN$ and $AO=NO$, we obtain for all $P$ on $x$- axis and for all $Q\in CD$:
$AP=NP$, $NQ=BQ$, $AP'=NP'$ and $NQ'=BQ'$.
Thus, 
$$AP+PQ+QB=MP+PQ+QN\geq MP'+P'Q'+Q'N=MN=\sqrt{1^2+6^2}=\sqrt{37}.$$
The equality occurs for $P\equiv P'$ and $Q\equiv Q'$, which says that $\sqrt{37}$ is the answer.
This problem we can prove also by the Minkowski's inequality.
Indeed, let $P(a,0)$ and $Q(b,3)$.
Thus, by Minkowski we obtain:
$$AP+PQ+QB=\sqrt{a^2+1}+\sqrt{(a-b)^2+9}+\sqrt{(b-1)^2+4}=$$ 
$$=\sqrt{a^2+1}+\sqrt{(b-a)^2+9}+\sqrt{(1-b)^2+4}\geq$$
$$\geq\sqrt{(a+b-a+1-b)^2+(1+3+2)^2}=\sqrt{37}.$$
Done!
A: The logic is:- "the shortest distance between two points is the straight line segment between them".
1) (Translate A to its equivalent position.) Find A' = (0, -1), the reflected image of A about the x-axis.
2) (Translate B to its equivalent position.) Find B' = (1, 5), the reflected image of B about the line y = 3.
3) Join A'B', the required line segment.
4) The place that A'B' cuts the x-axis is P. Q is similarly found. 
A: $P\left(\dfrac{1}{6};\;0\right)$ and $Q\left(\dfrac{2}{3};\;3\right)$
The graph is self explanatory. The shortest path is the straight line. I just reflected the given segments in a way such that the path gave a straight line and then reflected again to get back the points on the given segments. Points $M$ and $Q$ are on the same vertical line. 
Shortest distance is $r+m+n=\sqrt{1+\dfrac{1}{36}}+\sqrt{\dfrac{1}{4}+9}+\sqrt{4+\dfrac{1}{9}}=\sqrt{37}$

