Question

Prove that the equation
$y^3-x^3+3xy(y-x)=0$
represents three straight lines equally inclined to one another.

Answer 1

$y^3-x^3+3xy(y-x)=0$

Factorising the given equation

$(y-x)(y^2+x^2+xy)+3xy(y-x)=0(y-x)(y^2+x^2+xy+3xy)=0(y-x)(y^2+4xy+x^2)=0$

$y-x=\mathrm{0....}\left(i\right)y^2+4xy+x^2=0y=\frac{-4x\pm \sqrt{16x^2-4x^2}}{2}y=-2x\pm \sqrt{3}xy=-2x+\sqrt{3}x....\left(ii\right)y=-2x-\sqrt{3}x.....\left(iii\right)$

Angle between $\left(i\right)$ and $\left(ii\right)$

$\tan \theta =\frac{1-2+\sqrt{3}}{1-2+\sqrt{3}}=1\Rightarrow \theta ={45}^o$

Angle between $\left(ii\right)$ and $\left(iii\right)$

$\tan \theta =\frac{1-2-\sqrt{3}}{1-2-\sqrt{3}}=1\Rightarrow \theta ={45}^o$

Angle between $\left(ii\right)$ and $\left(iii\right)$

$\tan \theta =\frac{-2+\sqrt{3}-(-2-\sqrt{3})}{1+(-2+\sqrt{3})(-2-\sqrt{3})}=\frac{2\sqrt{3}}{0}=\infty \Rightarrow \theta ={90}^o$

Hence, they are equally inclined to each other.