A) \[x+2y=-\frac{1}{3}\]
B) \[x-2y=1\]
C) \[x+3y=-\frac{2}{3}\]
D) \[2x+y=\frac{1}{3}\]
Correct Answer: B
Solution :
[b]:\[2x-y=1\Rightarrow x=\alpha ,y=2\alpha -1\] The image of the point \[(\alpha ,2\alpha -1)\]in the line \[x+y=0\]is given by \[\frac{x-\alpha }{1}=\frac{y-(2\alpha -1)}{1}=-2\frac{(\alpha +2\alpha -1)}{1+1}=1-3\alpha \] \[\therefore \]\[x=1-2\alpha ,y=2\alpha -1+1-3\alpha =-\alpha \] Elimination of\[\alpha \]gives the image \[x-2y=1\].
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