| 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 |
| [SSC (CGL) 2011] |
A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) 2
D) \[-\,2\]
Correct Answer: A
Solution :
| \[(k-1)x+y-2=0\] |
| and \[(2-k)\,x-3y+1=0\] |
| \[{{m}_{1}}=1-k\] |
| and \[{{m}_{2}}=\frac{(2-k)}{3}\] |
| Since, the two times are parallel. |
| \[\therefore \] \[{{m}_{1}}={{m}_{2}}\] |
| \[\Rightarrow \] \[1-k=\frac{2-k}{3}\] |
| \[\Rightarrow \] \[3-3k=2-k\] |
| \[\Rightarrow \] \[-\,2k=-1\]\[\Rightarrow \]\[k=\frac{1}{2}\] |
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