A) \[(1,2)\]
B) \[(2,1)\]
C) \[(2,2)\]
D) \[(1,1)\]
Correct Answer: D
Solution :
| [d] Let \[P(x,y)\] be the point equidistant from the points \[A(0,0),\,\,\,B(2,0)\]and \[C(0,2)\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,PA=PB\] |
| \[\Rightarrow \,\,\,\,\,\sqrt{{{(x-0)}^{2}}+{{(y-0)}^{2}}}=\sqrt{{{(x-2)}^{2}}+{{(y-0)}^{2}}}\] |
| Squaring both sides, we get |
| \[{{x}^{2}}+{{y}^{2}}={{x}^{2}}+4-4x+{{y}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4x=4\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=1\] |
| Also, \[PA=PC\] |
| \[\Rightarrow \,\,\,\sqrt{{{(x-0)}^{2}}+{{(y-0)}^{2}}}=\sqrt{{{(x-0)}^{2}}+{{(y-2)}^{2}}}\] |
| Squaring both sides, we get, |
| \[{{x}^{2}}+{{y}^{2}}={{x}^{2}}+{{y}^{2}}+4-4y\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4y=4\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=1\] |
| Hence, the required point is \[(1,1)\] |
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