Question

Show that the following equation represent a pair of lines, find the acute angle between them
$2x^2-xy-3y^2-6x+19y-20=0$

Answer 1

Given equation of straight line is $2x^2-xy-3y^2-6x+19y-20=0$.

Compairing with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, we get

$a=2,h=-\frac{1}{2},b=-3,g=-3,f=\frac{19}{2},c=-20$

If the equation represents a pair of lines then,

$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$

$D=\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=\left|\begin{array}{ccc}2& \frac{-1}{2}& -3\\ \frac{-1}{2}& -3& \frac{19}{2}\\ -3& \frac{19}{2}& -20\end{array}\right|$

$D=2(60-\frac{361}{4})-\left(\frac{-1}{2}\right)(10+\frac{57}{2})+(-3)(\frac{-19}{4}-9)$

$D=2\times \frac{240-361}{4}+\frac{1}{2}\times \frac{20+57}{2}-3\times \frac{-19-36}{4}$

$D=\frac{-121}{2}+\frac{77}{4}+\frac{165}{4}$

$D=\frac{-242+242}{4}=0$

Since, determinant of the given equation is zero, it represents a pair of lines.

Let $\theta$ be the acute angle between the pair lines.

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

$\tan \theta =\left|\frac{2\sqrt{{\left(\frac{1}{2}\right)}^2-2\times (-3)}}{2+(-3)}\right|$

$\tan \theta =\left|\frac{2\sqrt{\frac{1}{4}+6}}{-1}\right|$

$\tan \theta =|-2\times \frac{5}{2}|$

$\therefore$ $\tan \theta =5$

$\therefore$ $\theta ={\tan }^{-1}\left(5\right)={78.69}^o$