A) \[2,3\]
B) \[-2,4\]
C) \[1,5\]
D) \[5,-1\]
Correct Answer: C
Solution :
| [c]\[AP=BP\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A{{P}^{2}}=B{{P}^{2}}\](By distance formula) |
| \[\therefore \,\,{{(k-1-3)}^{2}}+{{(2-k)}^{2}}={{(k-1-k)}^{2}}+{{(2-5)}^{2}}\] |
| \[\Rightarrow \,\,{{(k-4)}^{2}}+{{(2-k)}^{2}}={{(-1)}^{2}}+{{(-3)}^{2}}\] |
| \[\Rightarrow \,\,{{k}^{2}}+16-8k+4+{{k}^{2}}-4k=1+9\] |
| \[\Rightarrow \,\,2{{k}^{2}}-12k+10=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{k}^{2}}-6k+5=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{k}^{2}}-5k-k+5=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(k-5)\,\,(k-1)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k=1,5\] |
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