A) \[k=12\]
B) \[k=-12\]
C) \[k=9\]
D) \[k=-9\]
Correct Answer: C
Solution :
[c] The given equations are |
\[x+3y-4=0\] and \[3x+ky+12=0\] |
Here, \[{{a}_{1}}=1,\,{{b}_{1}}=3,\,{{c}_{1}}=-4,\,{{a}_{2}}=3,{{b}_{2}}=k,\,{{c}_{2}}=12\] |
For inconsistent equations: |
\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\Rightarrow \frac{1}{3}=\frac{3}{k}\ne \frac{-4}{12}\] |
\[\Rightarrow \,\,\,\,\,\,k=9\]and \[k\ne -9\] |
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