CBSE BoardEnglish MediumSTD 11 ScienceMathsStraight Lines2 Marks
Question
Find the angle between the lines $y-\sqrt{3} x-5=0$ and $\sqrt{3} y-x+6=0$.
✓
Answer
Given lines are
$y-\sqrt{3} x-5=0 \text { or } y=\sqrt{3} x+5$ .........(1)
and $\sqrt{3} y-x+6=0 \text { or } y=\frac{1}{\sqrt{3}} x-2 \sqrt{3}$......(2)
Slope of line (1) is $m_{1}=\sqrt{3}$ and slope of line (2) is $m_{2}=\frac{1}{\sqrt{3}}$
The acute angle (say) $\theta$ between two lines is given by
$\tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|$......(3)
Substituting,the values of $m_1$ and $m_2$ in (3), we obtain
$\tan \theta=\left|\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}\right|=\left|\frac{1-3}{2 \sqrt{3}}\right|=\frac{1}{\sqrt{3}}$
which gives $\theta = 30^\circ .$
Therefore,the angle between two lines is either $30^\circ$ or 1$80^\circ – 30^\circ = 150^\circ .$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.