A) 2
B) 1
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
E) \[-2\]
Correct Answer: E
Solution :
Area of triangle \[=\frac{1}{2}\left| \begin{matrix} k & 2k & 1 \\ 2k & 3k & 1 \\ 3 & 1 & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} k & 2k & 1 \\ k & k & 1 \\ 3-k & 1-2k & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[1[k(1-2k)-k(2-k)]=0\] \[\Rightarrow \] \[k-2{{k}^{2}}-3k+{{k}^{2}}=0\] \[\Rightarrow \] \[-{{k}^{2}}-2k=0\] \[\Rightarrow \] \[-k(k+2)=0\Rightarrow k=0,-2\]
You need to login to perform this action.
You will be redirected in
3 sec
