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 \[\left( \frac{5}{2},\,0 \right)\]You need to login to perform this action.
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