Question

The angle between the pair of straight lines $3x^2+4xy-3y^2=0$ is

• A. ${60}^0$
• B. ${90}^0$
• C. ${\tan }^{-1}\left(\frac{3}{4}\right)$
• D. ${\tan }^{-1}\left(\frac{4}{3}\right)$

Answer 1

Answer: (B)

${90}^0$

Given : Equation of the pair of straight lines

$3x^2+4xy-3y^2=0$.

We know that the standard equation of the pair of straight lines is

$a{x}^2+2hxy+b{y}^2=0$.

Comparing the given equation with the standard equation, we get

$a=3,h=2$ and $b=-3$

We also know that the relation for the angle between the pair of straight line $\left(\theta \right)$ is

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{2\sqrt{{\left(2\right)}^2-\left(3\right)\times (-3)}}{3+(-3)}=\infty$

$\Rightarrow \theta =\tan {}^{-1}=\left(\infty \right)={90}^0$