Question

Show that the equation $x^2-6xy+5y^2+10x-14y+9=0$ represent a pair of lines, Find the acute angle between them.

Answer 1

Given equation of straight line is $x^2-6xy+5y^2+10x-14y+9=0$.

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

$a=1,h=-3,b=5,g=5,f=-7,c=9$

If the equation represents a pair of lines then,

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

$D=\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=\left|\begin{array}{ccc}1& -3& 5\\ -3& 5& -7\\ 5& -7& 9\end{array}\right|$

$D=1(45-49)-(-3)(-27+35)+\left(5\right)(21-25)$

$D=-4+24-20$

$\therefore$ $D=0$

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

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

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

$\tan \theta =\left|\frac{2\sqrt{(-3{)}^2-1\times 5}}{1+5}\right|$

$\tan \theta =\left|\frac{2\sqrt{9-5}}{6}\right|=2\frac{\sqrt{4}}{6}$

$\tan \theta =|2\times \frac{2}{6}|$

$\therefore$ $\tan \theta =\frac{2}{3}$

$\therefore$ $\theta ={\tan }^{-1}\left(\frac{2}{3}\right)={33.69}^o$