Strait Line

\hspace{-16}$Here $\mathbf{x^3+ax^2y+bxy^2+y^3=0}$\\\\\\ Divide both side by $\mathbf{x^3}$, Where $\mathbf{x\neq 0}$\\\\\\ $\mathbf{1+a\left(\frac{y}{x}\right)+b\left(\frac{y}{x}\right)^2+\left(\frac{y}{x}\right)^3=0}$\\\\\\ Now Let $\mathbf{\left(\frac{y}{x}\right)=t}$, Then \\\\\\ $\mathbf{t^3+bt^2+at+1=0=(t-m_{1}.x)(t-m_{2}.x)(t-m_{3}.x)}$\\\\\\ Now Camparing Coefficient of Const., We Get\\\\\\ $\mathbf{m_{1}.m_{2}.m_{3}=-1}$\\\\\\ But Given That $\mathbf{m_{1}.m_{2}=-1}$\\\\\\ So $\mathbf{m_{3}=1}$\\\\\\ So Third equation is $\mathbf{(t-m_{3}.x)=\frac{y}{x}-1=0}$\\\\\\ So Third equation is $\boxed{\boxed{\mathbf{(y=x)}}}$

Correct Solution. I had found the same. But I found that two lines are then co incident i.e, the solutions are y=x,y=-x,y=x !

So I thought my solution is incorrect! :P