A) 2, 3
B) 1, 0
C) 1, 2
D) \[1,-1/2\]
E) 0, 3
Correct Answer: E
Solution :
\[\because \]Points\[(k,3),(2,k),(-k,3)\]are collinear. \[\left| \begin{matrix} k & 3 & 1 \\ 2 & k & 1 \\ -k & 3 & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[k(k-3)-3(2+k)+1(6+{{k}^{2}})=0\] \[\Rightarrow \] \[{{k}^{2}}-3k-6-3k+6+{{k}^{2}}=0\] \[\Rightarrow \] \[2{{k}^{2}}-6k=0\] \[\Rightarrow \] \[k(k-3)=0\] \[\Rightarrow \] \[k=0,3\]You need to login to perform this action.
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