Q78 / 180 · COMEDK 2014 · Mathematics

The angle between lines $\sqrt{3} x+y=1$ and $x+\sqrt{3} y=1$ is

  • Correct answer
    A
    $\frac{\pi}{6}$
  • B
    $\frac{3 \pi}{4}$
  • C
    $\frac{5 \pi}{2}$
  • D
    $\frac{\pi}{3}$

Solution

Given, equations of lines are $$ \sqrt{3} x+y=1 \quad \text{...(i)} $$ and $$ x+\sqrt{3} y=1 \quad \text{...(ii)} $$ Let $m_{1}$ and $m_{2}$ be slopes of Eqs. (i) and (ii), then $$ m_{1}=-\sqrt{3}, m_{2}=-\frac{1}{\sqrt{3}} $$ Let $\theta$ be angle between them, then $$ \begin{aligned} & \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|=\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+\sqrt{3} \cdot \frac{1}{\sqrt{3}}}\right|=\left|\frac{-3+1}{2 \sqrt{3}}\right|=\frac{1}{\sqrt{3}} \\ \Rightarrow & \theta=\frac{\pi}{6} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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