A) \[(x+a)(x+ay)=cy\]
B) \[(x+a)(1-ay)=cy\]
C) \[(x+a)(1-ay)=c\]
D) None of the above
Correct Answer: B
Solution :
We have the equation \[\left| \begin{matrix} x-a & x-b & x-c \\ x-b & x-c & x-a \\ x-c & x-a & x-b \\ \end{matrix} \right|=0\] Applying the operation\[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}},\]we get \[\left| \begin{matrix} 3x-(a+b+c) & x-b & x-c \\ 3x-(a+b+c) & x-c & x-a \\ 3x-(a+b+c) & x-a & x-b \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[[3x-(a+b+c)]\left| \begin{matrix} 1 & x-b & x-c \\ 1 & x-c & x-a \\ 1 & x-a & x-b \\ \end{matrix} \right|=0\] Applying\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\]and\[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\]we get \[[3x-(a+b+c)]\left| \begin{matrix} 1 & x-b & x-c \\ 0 & b-c & c-a \\ 0 & b-a & x-b \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\{3x-(a+b+c)\}\{(b-c)(c-b)\] \[-(b-a)(c-a)\}=0\] \[\Rightarrow \]\[\{3x-(a+b+c)\}\{-{{b}^{2}}-{{c}^{2}}+2bc-bc\] \[-bc-ca)\}=0\] \[\Rightarrow \]\[\{3x-(a+b+c)\}\{-({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab\] \[-bc-ca)\}=0\] \[\Rightarrow \]\[\{3x-(a+b+c)\}(2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab\] \[-2bc-2ca)=0\] \[\Rightarrow \]\[\{3x-(a+b+c)\}\{{{(a-b)}^{2}}+{{(b-c)}^{2}}\] \[+{{(c-a)}^{2}}\}=0\] \[\Rightarrow \]\[3x-(a+b+c)=0\] \[[\therefore {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}\ne 0\]as a, b, c are different] \[\Rightarrow \] \[3x=a+b+c\] \[x=\frac{1}{3}(a+b+c)\]
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