Question

The angle between the lines represented by the equation $2x^2+3xy-5y^2=0$, is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2}$
• C. ${\tan }^{-1}\left(\frac{12}{5}\right)$
• D. ${\tan }^{-1}(-\frac{7}{3})$

Answer 1

Answer: (D)

${\tan }^{-1}(-\frac{7}{3})$

Given equation is
$2x^2+3xy-5y^2=0$

On comparing with
$a{x}^2+2hxy+b{y}^2=0$, we get

$a=2,h=\frac{3}{2},b=-5$

$\therefore \tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$

$=\frac{2\sqrt{{\left(\frac{3}{2}\right)}^2+10}}{2-5}$

$=\frac{2\sqrt{\frac{9}{4}+10}}{-3}$

$=\frac{\sqrt{49}}{-3}$

$\Rightarrow \theta ={\tan }^{-1}(-\frac{7}{3})$