A) \[\left( \frac{5}{3},\,0 \right)\]
B) \[\left( \frac{1}{3},\,0 \right)\]
C) (3, 0)
D) (1, 0)
Correct Answer: A
Solution :
Let co-ordinate of R (x, 0) Given \[P(1,\,1)\] and \[Q\,(3,\,2)\] \[PR+RQ=\sqrt{{{(x-1)}^{2}}+{{(0-1)}^{2}}}+\sqrt{{{(x-3)}^{2}}+{{(0-2)}^{2}}}\] = \[\sqrt{{{x}^{2}}-2x+2}+\sqrt{{{x}^{2}}-6x+13}\] For minimum value of PR + RQ, \[\frac{d}{dx}(PR+RQ)=0\] Þ \[\frac{d}{dx}(\sqrt{{{x}^{2}}-2x+2})+\frac{d}{dx}(\sqrt{{{x}^{2}}-6x+13})=0\] Þ \[\frac{(x-1)}{\sqrt{{{x}^{2}}-2x+2}}=-\frac{(x-3)}{\sqrt{{{x}^{2}}-6x+13}}\] Squaring both sides, \[\frac{{{(x-1)}^{2}}}{({{x}^{2}}-2x+2)}=\frac{{{(x-3)}^{2}}}{{{x}^{2}}-6x+13}\] Þ\[3{{x}^{2}}-2x-5=0\]Þ \[(3x-5)\,(x+1)=0\], \[x=\frac{5}{3},\,-1\]. Also \[1<x<3\]. \[\therefore \] \[R=(5/3,\,0)\].
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