A) \[\frac{6}{19}\]
B) \[\frac{19}{6}\]
C) \[-\frac{19}{6}\]
D) \[-\frac{6}{19}\]
Correct Answer: B
Solution :
Given lines are, \[x+3y-6=0\] \[2x+y-4=0\] \[kx-3y+1=0\] these given lines will concurrent when \[\left| \begin{matrix} 1 & 3 & -\,6 \\ 2 & 1 & -\,4 \\ k & -\,3 & 1 \\ \end{matrix} \right|=0\] Expand with respect to \[{{R}_{1}}:\] \[1(1-12)-3(2+4k)-6\,(-6-k)=0\] \[\Rightarrow \] \[-11-6-12k+36+6k=0\] \[\Rightarrow \] \[=-6k-17+36=0\] \[\Rightarrow \] \[6k=19\] \[\Rightarrow \] \[k=\frac{19}{6}\]You need to login to perform this action.
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