A light beam from the point A(3,10) reflects from the straight line $2\mathrm{x}+\mathrm{y}-6=0$  and then passes through the point B(7,2) . Find the equations of the incident and reflected beams. 

detailed solution

Correct option is A

The incident ray will passes through the point  A(3,10) and the image of B(7,12) with respect to the line 2x+y−6=0 Suppose that the image of  B(7,2) with respect  to the line 2x+y−6=0  is P(h,k)It implies h−72=k−21=−2(2(7)+2−6)22+12 ∵h−x1a=k−y1b=−2ax1+by1+ca2+b2h−72=k−21=−2(10)5h−72=k−21=−4 Hence,h−72    =−4      k−21=−4h−7    =−8   and  k−2=−4h=−1                   k=−2  The image is   (-1,2)Equation of incident line is the equation of the line joining  A(3,10) and P(−1,−2)It implies y−10=−2−10−1−3(x−3)y−10=3(x−3)y−10=3x−93x−y+1=0Therefore, the equation of the incident ray is  3x−y+1=0The reflected ray is passing through the point of intersection of the incident ray and the surface lineThe point of intersection of 3x−y+1=0  and 2x+y−6=0 is  (1,4)Hence the equation of the reflected ray is equation of the line joining  (1,4) and  (7,2)It implies y−4=2−47−1(x−1)y−4=−13(x−1)3y−12=−x+1x+3y−13=0Therefore, the equation of the reflected ray is  x+3y−13=0