If the equation of the mirror be 2x + y – 6 = 0 and a ray passing through (3, 10) after being reflected by the mirror passes through (7, 2), then the equations of the incident ray and the reflected ray are

Equation of a straight line passing through the point (7, 2) be
y – 2 = m(x – 7)                                                                              (1)
and that of a straight line passing through the point (3, 10) be
y – 10 = m′ (x – 3)                                                                           (2)

Since the angle made by the incident and the reflected rays with the mirror must be same, therefore we have

$\frac{-2-m\text{'}}{1-2m\text{'}}=\frac{m-(-2)}{1-2m}$

i.e., 3m + 3m′ + 4mm′ – 4 = 0                                                            (3)
Also, since the lines (1), (2) and the mirror will be concurrent, therefore we have

$\left|\begin{array}{ccc}m& -1& 2-7m\\ m\text{'}& -1& 10-3m\text{'}\\ 2& 1& -6\end{array}\right|=0$

i.e., 18m – 2m′ + 4mm′ + 16 = 0                                                      (4)
Subtracting equation (4) from equation (3), we get
m′ = 3m + 4                                                                                      (5)
Now, solving equations (3), (5) we have

$m=-2.\frac{1}{3} and m\text{'}=-2, 3$

Since m = m′ = –2 is not acceptable, therefore equations of the required lines are

$y-2=-\frac{1}{3}(x-7) and y-10=3(x-3)$