Question

If incident ray passes through point $A(3,1)$ and reflected ray passes through point $B(5,6)$ and the plane mirror is kept at line $x + 2y - 1 = 0$ then, find the equation of incident and reflected ray.

Answer 1

Hint : In order to solve this question, we will first write down the general equation of line passing through a point and having slope m, for bothy rays of incident and reflected and then will find a common solution of these three equations of incident ray, reflected ray and mirror line to find values of slope of rays and then will find the equation of incident and reflected ray.
Formula used: If a line passes through a point (a, b) and having slope m then general equation can be written in the form of $\dfrac{{(y - b)}}{{(x - a)}} = m$

Complete step-by-step answer :
Let us suppose that, incident ray has a slope of m and reflected ray have a slope of m’ then according to slope definition, the equation of incident ray can be written while passing through given point $A(3,1)$ as
$\dfrac{{y - 1}}{{x - 3}} = m$
$y = (x - 3)m + 1 \to (i)$
Similarly for the reflected ray passing through a point $B(5,6)$ have the equation as
$\dfrac{{y - 6}}{{x - 5}} = m'$
$y = (x - 5)m' + 6 \to (ii)$
And given the equation of mirror line as $x + 2y - 1 = 0$ or can be written as
$2y = 1 - x \to (iii)$
Since three lines altogether will meet at a point where incident ray strike on mirror and reflected ray reflects from that point which means all three lines have a unique solution which can be found as
Adding the equation of (i) and (ii) and then compare it with equation third for common value of (x, y)
$(i) + (ii)$
$2y = x(m + m') - 3m - 5m' + 7$
Compare above equation with equation (iii) we get,
$2y = x(m + m') - 3m - 5m + 7$ And equation $2y = 1 - x$
Coefficient of x in both equation can be compared as
$m + m' = - 1$
And constant terms as
$3m + 5m' = 6$
So solving above two equations for m and m’ we have,
$3m + 5( - 1 - m) = 6$
$- 2m = 11$
$\Rightarrow m = \dfrac{{ - 11}}{2}$
And put this value in equation $m + m' = - 1$ we get,
$m' = \dfrac{9}{2}$
Now, for incident ray slope of line $y = (x - 3)m + 1 \to (i)$ put the value of $m = \dfrac{{ - 11}}{2}$ we get,
$2y = (x - 3)( - 11) + 2$
$\Rightarrow 2y = 35 - 11x$ Which is the equation of incident ray.
For reflected ray $y = (x - 5)m' + 6 \to (ii)$ put the value of $m' = \dfrac{9}{2}$ we get,
$2y = (x - 5)9 + 12$
$\Rightarrow 2y = 9x - 33$ Which is the equation of reflected ray.

Note : It should be remembered that, incident ray, reflected ray and the line of mirror will have the same solution at the point of intersection of these three lines and hence the simultaneous solution is used to find out the slopes of incident ray and reflected ray, physically slope is the measurement of a line how much the tangent of the angle line make with positive X axes.