Q: A ray of light generated from the source kept at (–3, 4) strikes the line 2x + y = 7 at R and then terminated at (0, 1). Find the point R so that ray travels through the shortest distance.

Sol. Slope of the line 2x + y = 7 is –2.

Let R be (a, 7 – 2a ), P (–3, 4) and Q (0, 1)

Slope of PR $\displaystyle = \frac{3-2a}{a+3}$ ,

Slope of RQ $\displaystyle = \frac{6-2a}{a}$

Now, ray travels through the shortest distance so, PR must be incident ray and RQ must be reflected ray.

$\displaystyle \frac{\frac{3-2a}{a+3} + 2}{1-2(\frac{3-2a}{a+3})} = \frac{-2 -\frac{6-2a}{a} }{1-2(\frac{6-2a}{a})}$

$\displaystyle a = \frac{42}{25} $

Hence R is $\displaystyle ( \frac{42}{25} , \frac{91}{25}) $