A) 2
B) 1
C) -1
D) 0
Correct Answer: D
Solution :
, Given, \[|a-c|\,=\,|b-c|\] \[\Rightarrow \] \[|a-c{{|}^{2}}=\,|b-c{{|}^{2}}\] \[|a{{|}^{2}}+|c{{|}^{2}}-2a\cdot c=|b{{|}^{2}}+|c{{|}^{2}}-2b\cdot c\] \[\Rightarrow \] \[2b\cdot c-(|b{{|}^{2}}-|c{{|}^{2}})-2a\cdot c=0\] \[\Rightarrow \] \[b\cdot c-\frac{1}{2}\,(|b{{|}^{2}}-|c{{|}^{2}})-a\cdot c=0\] ?(i) \[\therefore \] \[(b-a)\cdot \,\left( c-\frac{a+b}{2} \right)=b\cdot c-b\cdot \,\left( \frac{a+b}{2} \right)\] \[-a\cdot c+\frac{a}{2}\,\cdot \,(a+b)\] \[=b\cdot c-\frac{1}{2}\,(|b{{|}^{2}}-|a{{|}^{2}})-a\cdot c\] [from Eq. (i)] = 0
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