A) \[3x+4y=14\]
B) \[8x+6y=28\]
C) \[12x+9y=42\]
D) \[-12x=9y\]
Correct Answer: D
Solution :
| [d] \[4x+3y=14\] |
| Considering \[12x+9y=0\] |
| \[{{a}_{1}}=12,\,{{b}_{1}}=9,\,\,{{c}_{1}}=0\] |
| \[{{a}_{2}}=4,\,{{b}_{2}}=3,\,\,{{c}_{2}}=14\] |
| \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{12}{4}=3,\,\,\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{9}{3}=3\] |
| \[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{0}{14}=0\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\] |
| \[\therefore \] Lines are parallel. |
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