A) 1
B) \[-1\]
C) 25
D) 5
E) \[-20\]
Correct Answer: C
Solution :
\[\because \]The given lines are concurrent, then \[\left| \begin{matrix} 1 & -1 & -1 \\ 4 & 3 & -k \\ 2 & -3 & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[1\left| \begin{matrix} 3 & -k \\ -3 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} 4 & -k \\ 2 & 1 \\ \end{matrix} \right|-1\left| \begin{matrix} 4 & 3 \\ 2 & -3 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[1(3-3k)+1(4+2k)-1(12-6)=0\] \[\Rightarrow \] \[k=25\]You need to login to perform this action.
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