Question

Prove that the following equation represent two straight lines; find also their point of intersection and the angle between them.
$x^2-5xy+4y^2+x+2y-2=0.$

Answer 1

The general eqaution of second degree $a{x}^2+2hxy+{by}^2+2gx+2fy+c=0$ Represents a pair of straight lines if $\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

For $x^2-5xy+4y^2+x+2y-2=0$

$a=1,b=4,h=\frac{-5}{2},g=\frac{1}{2},f=1,c=-2\Delta =(1\times 4\times -2)+(2\times 1\times \frac{1}{2}\times \frac{-5}{2})-(1\times 1\times 1)-(4\times \frac{1}{2}\times \frac{1}{2})-(-2\times \frac{-5}{2}\times \frac{-5}{2})\Delta =-8-\frac{5}{2}-1-1+\frac{25}{2}=0$

$\therefore$ The equation represents a pair of straight lines. The point of intersection is found by partially differentiating the equation first with respect to $x$ and then with respect to $y$ and solving both the equations

$\frac{\partial }{\partial x}(x^2-5xy+4y^2+x+2y-2=0)2x-5y+0+1+0+0=02x-5y+1=\mathrm{0....}\left(i\right)\frac{\partial }{\partial y}(x^2-5xy+4y^2+x+2y-2=0)0-5x+0+8y+0+2-0=0-5x+8y+2=\mathrm{0...}\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$

we get $x=2,y=1$

So the point of intersection is $(2,1)$

Angle betwwen apair of staright lines that is $\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

$\tan \theta =\left|\frac{\sqrt{{\left(\frac{-5}{2}\right)}^2-\left(1\right)\left(4\right)}}{1+4}\right|=\frac{2\left(\frac{3}{2}\right)}{5}=\frac{3}{5}\theta ={\tan }^{-1}\left(\frac{3}{5}\right)$