question_answer

                A light ray emerging from the point source placed at P(1, 3) is reflected at a point Q in the axis of \[x\]. If the reflected ray passes through the point R(6,7), then the abscissa of is:                   JEE Main Online Paper (Held On 09 April 2013)

A)                 1                

B)                 3                

C)                 \[\frac{7}{2}\]                

D)                 \[\frac{5}{2}\]                

Correct Answer: D

Solution :

                Here, \[AS\bot OX\]                                 It means AS bisect the angle PAR. Then      \[PAS=RAS\] \[\Rightarrow \]               \[RAX=PA'=\theta \,(\text{let})\] \[\Rightarrow \]               \[XAP={{180}^{0}}-\theta \]                 Slope of \[AR=\tan \theta =\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{7-0}{6-k}\]            ?(i)                 Slope of \[AP=\tan \,({{180}^{0}}-\theta )=-\tan \theta \]                 \[=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{3-0}{1-k}\] \[\therefore \] From Eqs. (i) and (ii).                 \[\frac{7}{6-k}=-\frac{3}{1-k}\Rightarrow \,7-7k=-18+3k\]                 \[10k=25\,\,\Rightarrow \,\,k=\frac{5}{2}\]                 Hence, the coordinate of A is \[\left( \frac{5}{2},\,0 \right)\]