Find $R$ such that the man covers shortest distance. 
A man starts from $P(-3,4)$ and will reach the point $Q(0,1)$ touching the line $2x+y=7$ at $R$. Find $R$ on the line so that he will travel along the shortest distance.

I did not understand the question and after drawing the diagram , I found out that the line $2x+y=7$ does not touch the line joining $P(-3,4)$ and $Q(0,1)$.
Thanks
 A: HINT
R is that point on the line such that the angles of PR and QR with the line are equal.
A trick is to consider the reflection $\bar P$ of $P$ (or Q) with respect to the line and then consider the intersection between $\bar P Q$ and the line. The intersection point is R.

A: Let $Q'$ be symmetrical to $Q$ with respect to the line $l:y=-2x+7.$
Also, let $PQ'\cap l=\{R\}$.
Thus, $R$ is a needed point.
A: Refer to Gimusi's drawing:
Let me use $\overline{AC}: = $length $AC.$
Let $A'$ be the reflection of $A$ with respect to the given line. 
If $A',B,C$ are collinear then 
$\overline {AC}+ \overline {CB}$ is shortest.
Proof:
Line $A'B$ intersects the given line at $C.$
Choose any other path from $A$ to $B$,  
i.e a point $D \not = C$.
Consider the $\triangle A'DB$ .
The sum of 2 sides of a triangle is greater than the third side.
Hence: 
$\overline {A'D} +  \overline{DB} \gt \overline{ A'B}$, 
or, since $\overline{ A'D }= \overline{ AD}$ , and 
$\overline{ A'B} = \overline {AC} + \overline{CB}$, 
we have:
$\overline{AD} + \overline {DB} > \overline{AC} + \overline{CB}.$
