A) y-axis
B) a straight line
C) a circle
D) None of these
Correct Answer: C
Solution :
Since the lines \[x+2ay+a=0\] \[x+3by+b=0\] and \[x+4cy+c=0\] are concurrent \[\therefore \] \[\left| \begin{matrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \\ \end{matrix} \right|=0\] Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}},\]we get \[\left| \begin{matrix} 1 & 2a & a \\ 0 & 3b-2a & b-a \\ 0 & 4c-2a & c-a \\ \end{matrix} \right|=0\] \[(3b-2a)\text{ (}c-a)-(4c-2a)\text{ (}b-a)=0\] \[\Rightarrow \]\[-3bc-3ab-2ac+2{{a}^{2}}-4bc-4ac\] \[+2ab-2{{a}^{2}}=0\] \[\Rightarrow \] \[-bc-ab+\text{ }2ac=0\] \[\Rightarrow \] \[2ac=ab+bc\] \[\Rightarrow \] \[\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\] Hence, a, b, c are in HP.
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