Q194 / 270 · NDA 2017 (Phase 2) · Mathematics

The angle between the lines $x+y-3=0$ and $x-y+3=0$ is $\alpha$ and the acute angle between the lines $x-\sqrt{3} y+2 \sqrt{3}=0$ and $\sqrt{3} x-y+1=0$ is $\beta$. Which one of the following is correct?

  • A
    $\alpha=\beta$
  • Correct answer
    B
    $\alpha \gt \beta$
  • C
    $\alpha \lt \beta$
  • D
    $\alpha=2 \beta$

Solution

The slope of line $x+y-3=0$ is -1 The slope of line $x-y+3=0$ is 1 So, these are perpendicular lines and so angle between them is $90^{\circ}$. $\therefore \alpha=90^{\circ}$ The slope of line $x-\sqrt{3} y+2 \sqrt{3}=0$ is $m_1=\frac{1}{\sqrt{3}}$ The slope of line $\sqrt{3} x-y+1=0$ is $m_2=\sqrt{3}$. $\therefore \tan \beta=\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+(\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)}\right|=\left|\frac{2}{2 \sqrt{3}}\right|=\frac{1}{\sqrt{3}} \therefore \beta=30^{\circ}$ $\therefore \alpha \gt \beta$

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

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