 Hi and welcome to our session. Let us discuss the following question. The question says a ray of light passing through the point 1, 2 reflects on the x-axis at point A and the reflected ray passes through the point 5, 3. Find the coordinates of A. We know that if this line is a mirror line, then this line is normal to this line and angle of incidence is equal to angle of reflection. Now in this question a ray of light is passing through the point 1, 2. So this means this incident ray is passing through the point 1, 2 with this point B, P and this ray reflects on x-axis at point A and the reflected ray that is this ray passes through the point 5, 3. Let this point be Q which is having coordinate 5, 3. We have to find the coordinates of A. Since A lies on the x-axis therefore it coordinates out of the form K0. Let us name this line as A n. Now from this figure we know that A n is normal to O x angle of incidence is equal to angle of reflection that is angle P a n is equal to angle Q a n. Angle P a n is equal to angle Q a n therefore it implies angle P a o is equal to angle Q a x. Now let both these angles to be equal to theta. Now as angle P a o is equal to theta therefore angle P a x is equal to 180 degree minus theta. We know that slope of a line is given by n is equal to tan alpha where alpha is the angle made by the line with the positive direction of x-axis. Now line AQ makes angle theta with x-axis so slope of line AQ is equal to tan theta. Now A has coordinates K0 and Q has coordinates pi 3 therefore slope of line AQ is also equal to 3 minus 0 upon 5 minus K. So 3 minus 0 upon 5 minus K is equal to tan theta. Let us name this equation as equation number 1. Now we will find slope of line A P. Now A P makes angle 180 degree minus theta with positive direction of x-axis therefore slope of line A P is equal to tan 180 degree minus theta. Now A has coordinates K0 and P has coordinates 1, 2. So slope of line A P is also equal to 2 minus 0 upon 1 minus K. So we have 2 minus 0 upon 1 minus K is equal to minus tan theta. Let us name this equation as equation number 2. Now from 1 and 2 we have 3 upon 5 minus K is equal to minus 2 upon 1 minus K. Now this implies 3 minus 3 K is equal to minus tan plus 2 K. This implies minus 5 K is equal to minus 13 and this implies K is equal to 13 by 5. Hence the required coordinates of point A are 13 by 5 and 0. This is our required answer. So this completes the session. Bye and take care.