Answer:
Let the coordinates of point P be \[(2y,\text{ }y)\] Since, P is equidistant from Q and R \[\therefore PQ=PR\] \[\Rightarrow \sqrt{{{(2y-2)}^{2}}+{{(y+5)}^{2}}}=\sqrt{{{(2y+3)}^{2}}+{{(y-6)}^{2}}}\] \[\Rightarrow {{(2y-2)}^{2}}+{{(y+5)}^{2}}={{(2y+3)}^{2}}+{{(y-6)}^{2}}\] \[\Rightarrow \,4{{y}^{2}}+4-8y+{{y}^{2}}+25+10y=4{{y}^{2}}+9+12y+{{y}^{2}}+36-12y\] \[\Rightarrow 2y+29=45\] \[\Rightarrow 2y=45-29\] \[\Rightarrow y=\frac{16}{2}=8\] Hence, the co-ordinates of point P are (16, 8).
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