A) \[\text{x}+\text{3y}=0\]
B) \[\text{3x}+\text{y}=0\]
C) \[\text{x}+\text{2y}=0\]
D) \[\text{3x}+\text{2y}=0\]
Correct Answer: A
Solution :
| [a] Let the point be \[P(x,y),\]\[A(2,1)\]and \[B(1,-2).\]Since, P is equidistant from A and B. |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,AP=BP\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A{{P}^{2}}=B{{P}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(x-2)}^{2}}+{{(y-1)}^{2}}={{(x-1)}^{2}}+{{(y+2)}^{2}}\] |
| \[\Rightarrow \,\,\,\,{{x}^{2}}-4x+4+{{y}^{2}}-2y+1={{x}^{2}}-2x+1+{{y}^{2}}+4y+4\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,x+3y=0\] |
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