Question

The angle between the pair of straight lines $y^2{\sin }^2\theta -xy{\sin }^2\theta +x^2({\cos }^2\theta -1)=0$ is :

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

Answer 1

Answer: (B)

$\frac{\pi }{2}$

$y^2{\sin }^2\theta -xy{\sin }^2\theta +x^2({\cos }^2\theta -1)=0$

$y^2{\sin }^2\theta -xy{\sin }^2\theta -x^2{\sin }^2\theta =0$

we know that for $a{x}^2+b{y}^2+2hxy=0$ if $a+b=0$ then the pair of lines are perpendicular

Here $a=si{n}^2\theta ,b=-{\sin }^2\theta \Rightarrow a+b=0$

Hence, the lines are perpendicular.