A) \[x+2y=1\]
B) \[x+2y=2\]
C) \[x+2y=4\]
D) \[x+2y=3\]
Correct Answer: D
Solution :
The intersection of two line is \[(x+y-2)+\lambda (x-y)=0\] or \[(1+\lambda )x+(1-\lambda )y-2=0\] Which is parallel to \[x+2y=5\] \[\therefore \] \[-\frac{(1+\lambda )}{(1-\lambda )}=-\frac{1}{2}\] \[\Rightarrow \] \[2+2\lambda =1-\lambda \] \[\Rightarrow \] \[\lambda =-\frac{1}{3}\] Put \[\lambda =-\frac{1}{3}\] in Eq. (i), we get \[\frac{2}{3}x+\frac{4}{3}y-2=0\] \[\Rightarrow \] \[x+2y-3=0\]
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