Question

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

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

Answer 1

The correct option is D

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

Find the angle between the lines

Consider the given equation;

$2x^2+3xy-5y^2=0......\left(1\right)$

On comparing the given equation of lines with the general equation $a{x}^2+2hxy+b{y}^2=0$, we get;

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

We know that, angle between pair of straight lines is expressed as,

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

Substitute the value $a=2,b=-5,\text{and}h=\frac{3}{2}$ in the above equation

$$\begin{gathered} \tan \theta =\left|\frac{2\sqrt{{\left({\displaystyle \frac{3}{2}}\right)}^2-\left(2\right)\left(-5\right)}}{2+\left(-5\right)}\right| \\ \tan \theta =\left|\frac{2\sqrt{{\displaystyle \frac{9}{4}}+10}}{2-5}\right| \\ \tan \theta =\left|\frac{2\sqrt{{\displaystyle \frac{49}{4}}}}{-3}\right| \\ \tan \theta =\left|\frac{2{\displaystyle \times \frac{7}{2}}}{-3}\right| \\ \tan \theta =\frac{7}{3} \\ \theta ={\tan }^{-1}\frac{7}{3} \end{gathered}$$

Hence, the correct answer is Option D.