A) 4
B) 0
C) -4
D) None of these
Correct Answer: C
Solution :
Let \[A\equiv (x+1,\,2),\,\,B\equiv (1,\,x+2),\,\,C\equiv \left( \frac{1}{x+1},\,\,\frac{2}{x+1} \right)\] then A, B, C are collinear if area of \[\Delta ABC=0\] \[\Rightarrow \,\,\left| \,\begin{matrix} x+1 & 2 & 1 \\ 1 & x+2 & 1 \\ \frac{1}{x+1} & \frac{2}{x+1} & 1 \\ \end{matrix}\, \right|=0\] \[\Rightarrow \,\,\left| \,\begin{matrix} x & -x & 0 \\ 1 & x+2 & 1 \\ \frac{1}{x+1} & \frac{2}{x+1} & 1 \\ \end{matrix}\, \right|=0\] \[({{R}_{1}}\to {{R}_{1}}-{{R}_{2}})\] \[\Rightarrow \,\,\left| \,\begin{matrix} x & 0 & 0 \\ 1 & x+3 & 1 \\ \frac{1}{x+1} & \frac{3}{x+1} & 1 \\ \end{matrix}\, \right|=0\] \[({{C}_{2}}\to {{C}_{2}}+{{C}_{1}})\] \[\Rightarrow \,\,x\,\left( x+3-\frac{3}{x+1} \right)=0\,\,\Rightarrow \,\,x({{x}^{2}}+3+4x-3)=0\] \[\Rightarrow \,\,{{x}^{2}}(x+4)=0\,\,\Rightarrow \,\,x=0,\,\,-4\].
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