Question

Find the equation of pair of lines through the origin each of which making an angle of ${30}^{\circ }$ with the line $3x+2y-11=0$.

Answer 1

• Equation of any line passing through origin is y=mx, where m is the slope of line.
• Let the slope of required pair of lines is $m_1=m$
• Slope of the given line is $m_2=-\frac{3}{2}$
• Given angle is $\theta ={30}^{\circ }$
• Substitute all the values in the formula $tan\theta =\mid \frac{m_1-m_2}{1+m_1{m}_2}\mid$

So, $tan{30}^{\circ }=\mid \frac{m+\frac{3}{2}}{1-\frac{3m}{2}}\mid$

$\text{or}$, $\pm \frac{1}{\sqrt{3}}=\frac{2m+3}{2-3m}$

$\text{or}$, $\pm (2-3m)=\sqrt{3}(2m+3)$

• $\text{Case 1 :}(2-3m)=\sqrt{3}(2m+3)$
  
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$\text{or}$, $m=\frac{2-3\sqrt{3}}{2\sqrt{3}+3}$

• $\text{Case 2:}-(2-3m)=\sqrt{3}(2m+3)$
  
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$\text{or}$, $m=\frac{3\sqrt{3}+2}{3-2\sqrt{3}}$


• Therefore, equation of required lines is
  
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$y=\left(\frac{2-3\sqrt{3}}{2\sqrt{3}+3}\right)x$ $\text{or}$ $y=\left(\frac{3\sqrt{3}+2}{3-2\sqrt{3}}\right)x$