Find angles between the lines
Question:

Find angles between the lines  $\sqrt{3} x+y=1$ and $x+\sqrt{3} y=1$

Solution:

The given lines are $\sqrt{3} x+y=1$ and $x+\sqrt{3} y=1$.

$y=-\sqrt{3} x+1$ $\ldots(1)$

and

$y=-\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}}$$\ldots(2)$

The slope of line (1) is $m_{1}=-\sqrt{3}$,

while the slope of line

(2) is $m_{2}=-\frac{1}{\sqrt{3}}$.

The acute angle i.e., $\theta$ between the two lines is given by

$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$

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

$\tan \theta=\left|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}\right|=\left|\frac{-2}{2 \times \sqrt{3}}\right|$

$\tan \theta=\frac{1}{\sqrt{3}}$

$\theta=30^{\circ}$

Thus, the angle between the given lines is either $30^{\circ}$ or $180^{\circ}-30^{\circ}=150^{\circ}$.

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