question_answer

The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from \[Q(2,-5)\]and \[R(-3,6)\], find the coordinates of P.

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).