Question

If the pair of lines $2x^2+3xy+y^2=0$ makes angles of ${\theta }_1$ and ${\theta }_2$ with the positive directions of the X-axis, then $|\tan ({\theta }_1-{\theta }_2)|$ =

• A. $1$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{1}{3}$

Comparing it with equation $a{x}^2+2hxy+b{y}^2=0$

$m_1+m_2=-\frac{2h}{b}=-\frac{3}{1}=-3m_1{m}_2=\frac{a}{b}=\frac{2}{1}=2$

$\tan ({\theta }_1-{\theta }_2)=\frac{tan{\theta }_1-tan{\theta }_2}{1+tan{\theta }_{1}tan{\theta }_2}=\frac{m_1-m_2}{1+m_1{m}_2}=\left|\frac{\sqrt{(m_1+m_2{)}^2-4m_1{m}_2}}{1+m_1{m}_2}\right|=\left|\frac{\sqrt{(-3{)}^2-4\times 2}}{1+2}\right|=\frac{1}{3}$