Question

Prove that the following equation represent two straight lines; find also their point of intersection and the angle between them.
$6y^2-xy-x^2+30y+36=0.$

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 $6y^2-xy-x^2+30y+36=0$

$a=-1,b=6,h=\frac{-1}{2},g=0,f=15,c=36\Delta =(-1\times 6\times 36)+(2\times \frac{-1}{2}\times 0\times -1)-(-1\times 15\times 15)-(6\times 0\times 0)-(36\times \frac{-1}{2}\times \frac{-1}{2})\Delta =-216+0+225-0-9\Delta =-216-216=0\Delta =0$

$\therefore$ 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.

$\frac{\partial }{\partial x}(6y^2-xy-x^2+30y+36=0)\Rightarrow 0-y-2x+0+0=0y+2x=\mathrm{0.....}\left(i\right)\frac{\partial }{\partial y}(6y^2-xy-x^2+30y+36=0)12y-x+0+30+0=012y-x+30=\mathrm{0....}\left(ii\right)$

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

we get $x=\frac{6}{5},y=\frac{-12}{5}$

So the point of intersection is $(\frac{6}{5},\frac{-12}{5})$

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

$\tan \theta =\left|\frac{\sqrt{{\left(\frac{-1}{2}\right)}^2-(-1)\left(6\right)}}{-1+6}\right|=\frac{5}{5}=1\theta ={\tan }^{-1}\left(1\right)={45}^o$