A) \[(16,8)\]
B) \[(14,7)\]
C) \[(18,9)\]
D) \[(10,5)\]
Correct Answer: A
Solution :
| [a] Let the coordinates of P be \[(x,y)\]. Since P is equidistant from \[Q(2,-5)\]and \[R(-3,6)\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,PQ=PR\] |
| \[\sqrt{{{(2-x)}^{2}}+{{(-5-y)}^{2}}}=\sqrt{{{(-3-x)}^{2}}+{{(6-y)}^{2}}}\] |
| Squaring both sides, we get |
| \[{{(2-2y)}^{2}}+{{(-5-y)}^{2}}={{(-3-2y)}^{2}}+{{(6-y)}^{2}}\] |
| [\[x=2y,\]Given] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,5{{y}^{2}}+2y+29=5{{y}^{2}}+45\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,2y=16\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,y=8\] |
| Hence, the coordinates of P are \[(16,8)\] |
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