| The value of k for which the graphs of \[(k-1)x+y-2=0\] and \[(2-k)\,x-3y+1=0\]are parallel, is |
A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) 2
D) \[-2\]
Correct Answer: A
Solution :
| The graphs of \[(k-1)\,x+y-2=0\] |
| and \[(2-k)\,x-3y+1=0\]are parallel. |
| Two straight lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] |
| and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] are parallel, if |
| \[\frac{{{a}_{1}}}{{{b}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}.\] |
| It means that the given system of equations has no solution. |
| \[\therefore \] \[\frac{k-1}{2-k}=\frac{1}{-3}\] |
| \[\Rightarrow \] \[-\,3k+3=2-k\] |
| \[\Rightarrow \] \[-\,3k+k=2-3\] |
| \[\Rightarrow \] \[-\,2k=-1\] |
| \[\therefore \] \[k=\frac{1}{2}\] |
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