A) \[y=\tan (\theta -{{30}^{o}})(x-4-3\sqrt{5})\]
B) \[y=\tan ({{30}^{o}}-\theta )(x-4-3\sqrt{5})\]
C) \[y=\tan (\theta +{{30}^{o}})(x+4+3\sqrt{5})\]
D) \[y=\tan (\theta -{{30}^{o}})(x+4+3\sqrt{5})\]
Correct Answer: A
Solution :
The equation of a line parallel to\[x+2y=4\]is\[x+2y=k\]. Since, the distance between these two lines is 3, therefore \[\frac{k}{\sqrt{1+4}}-\frac{4}{\sqrt{1+4}}=3\] \[\Rightarrow \] \[k=4+3\sqrt{5}\] This shifted line cuts x-axis at\[(k,\,\,0)\]. After rotation the slope of the line is\[(\theta -{{30}^{o}})\], where\[\tan \theta =\](slope of \[(x+2y=4)=-\frac{1}{2}\] \[\therefore \]The equation of the line in the new position is \[y-0=\tan (\theta -{{30}^{o}})(x-k)\] \[\Rightarrow \] \[y=\tan (\theta -{{30}^{o}})(x-k)\] where\[k=4+3\sqrt{5}\]
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