A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) 2
D) \[-\,2\]
Correct Answer: A
Solution :
[a] Given lines are \[(x-1)+x+y-2=0\] (i) and \[(2-k)x-3y+1=0\] (ii) Since, the lines (i) and (ii) are parallel. \[\therefore \] \[\frac{k-1}{2-k}=\frac{1}{-3}\] \[\Rightarrow \] \[3k-3=-\,2+k\] \[\Rightarrow \] \[2k=1\] \[\Rightarrow \] \[k=\frac{1}{2}\] |
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