Question

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) $\alpha =\beta$
• (B) $\alpha >\beta$
• (C) $\alpha <\beta$
• (D) $\alpha =2\beta$

Answer 1

The Correct Option is B

Explanation:
Given:Angle between the lines $x+y-3=0$ and $x-y+3=0$ is $\alpha$.Angle between the lines $x-\sqrt{3y}+2\sqrt{3}=0$ and $\sqrt{3x}-y+1=0$ is $\beta$.The angle $\theta$ between the lines having slope $m_1$ and $m_2$ is given by $\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$Let's find the slope,Slope of line $x+y-3=0$ is $m_1$ and slope of line $x-y+3=0$ is $m_2$$m_1=-1$ and $m_2=1$Therefore,$\tan \alpha =\left|\frac{1-(-1)}{1+(-1)\times 1}\right|=\mathrm{\infty }$$\alpha ={90}^{\circ }$Slope of line $x-\sqrt{3}y+2\sqrt{3}=0$ is $m_1$ and $\sqrt{3}x-y+1=0$ is $m_2$$m_1=\left(\frac{1}{\sqrt{3}}\right)$ and $m_2=\sqrt{3}$$\tan \beta =\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+(\sqrt{3}\times \frac{1}{\sqrt{3}})}\right|=\frac{1}{\sqrt{3}}$$\beta ={30}^{\circ }$$\alpha >\beta$Hence, the correct option is (B).