I've seen it before - 9

The shortest path is one that traveled by light starting from $P (-3,4)$ reflected by $x-axis$ , the mirror, at $R(\alpha, 0)$ and ending at $Q (0, 1)$ . Therefore the angle of incident $\theta_i =$ angle of reflection $\theta_r$ .

Therefore, $\quad \tan {\theta_i} = \tan {\theta_r}$

$\Rightarrow \dfrac {3-|\alpha|}{4} = \dfrac {|\alpha|}{1} \quad \Rightarrow 3-|\alpha| = 4|\alpha| \quad \Rightarrow 5|\alpha| = \boxed {3}$

This problem can also be solved by Euler principle of least action which originated from Fermat principle of least time which postulates that "light travels between two given points along the path of shortest time." (http://en.wikipedia.org/wiki/Principle of least_action)

The image of the Q as reflected by y=0, is (0, - 1). Line length ( - 3,4)-(0, - 1) is the shortest and intersect y=0 at $(\alpha,0).~ The~ line~ is~ y= \dfrac {4 - ( - 1)}{ - 3 - 0}x - 1,~ intersect~ y=0,~at~( -\dfrac 3 5,0).\\ \therefore~5|\alpha|=5*|-\dfrac 3 5|=3.$

Here we can use calculus, to minimize the sum of distances PR and RQ, first find out the sum of distances, that will be in terms of \alpha. now differentiate it once wrt \alpha and equate it to zero. you will find the two values of \alpha. one will be positive and other will be negative. for (PR+RQ) to be minimum the negative value will be favourable for the given condition.