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  4. The angle between the lines √3x-y-2=0 and x-√3y+1=0 is:

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Q. The angle between the lines $ \sqrt{3}x-y-2=0 $ and $ x-\sqrt{3}y+1=0 $ is:

KEAMKEAM 2006

Solution:

Let $ {{m}_{1}}= $ slope of $ \sqrt{3}x-y-2=0 $ i.e., $ {{m}_{1}}=\sqrt{3} $ and $ {{m}_{2}}= $ slope of $ x-\sqrt{3}y+1=0 $ i.e., $ {{m}_{2}}=\frac{1}{\sqrt{3}} $ $ \therefore $ $ \tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+1} \right|=\left| \frac{3-1}{2.\sqrt{3}} \right| $ $ =\frac{1}{\sqrt{3}} $ $ \therefore $ $ \theta =30{}^\circ $