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Q : 9     Find angles between the lines  \sqrt{3}x+y=1  and   x+\sqrt{3}y=1.

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Given equation of lines are
  \sqrt{3}x+y=1  and   x+\sqrt{3}y=1

Slope of line \sqrt{3}x+y=1 is, m_1 = -\sqrt3

And 
Slope of line x+\sqrt{3}y=1  is , m_2 = -\frac{1}{\sqrt3}

Now, if  \theta is the angle between the lines
Then,

\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |

\tan \theta = \left | \frac{-\frac{1}{\sqrt3}-(-\sqrt3)}{1+(-\sqrt3).\left ( -\frac{1}{\sqrt3} \right )} \right | = \left | \frac{\frac{-1+3}{\sqrt3}}{1+1} \right |=| \frac{1}{\sqrt3}|

\tan \theta = \frac{1}{\sqrt3} \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \tan \theta = -\frac{1}{\sqrt3}

\theta = \frac{\pi}{6}=30\degree \ \ \ \ \ \ \ or \ \ \ \ \ \ \theta =\frac{5\pi}{6}=150\degree

Therefore, the angle between the lines is 30\degree \ and \ 150\degree

Posted by

Gautam harsolia

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