Strait Line

The eq x3+ax2y+bxy2+y3 represents 3 straight lines. if 2 of them r perpendicular then find the eqn of the 3rd line..

3 Answers

1708
man111 singh ·

\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 $\boxed{\boxed{\mathbf{(y=x)}}}$

71
Vivek @ Born this Way ·

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

7
Sigma ·

Thanx man111.

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