A) coincident
B) parallel
C) intersecting exactly at one point
D) perpendicular to each other
Correct Answer: B
Solution :
| [b] Here, \[{{a}_{1}}=6,\,\,{{b}_{1}}=-2,\,\,{{c}_{1}}=9\] |
| and \[{{a}_{2}}=3,\,\,{{b}_{2}}=-1,\,{{c}_{2}}=12\] |
| \[\therefore \frac{{{a}_{1}}}{{{a}_{2}}}=\frac{6}{3}=\frac{2}{1},\,\,\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{-2}{-1}=\frac{2}{1}\]and \[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{9}{12}=\frac{3}{4}\] |
| Hence, \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\] |
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