Prove that lines $y^3 - x^3 + 3xy(y -x) = 0$ are equally inclined to each other.

Question

Prove that lines $y^3 - x^3 + 3xy(y -x) = 0$ are equally inclined to each other.

I have to find this by using polar form of this equation.

---

Without using polar form I get,

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

$$\implies y = x \ or \ y^2 + 4xy + x^2 = 0$$

Solving second part,

$$y^2 + 4xy + x^2 = 0 \implies x= y(-2 \pm \sqrt{3})$$

Therefore the angles made by these three lines with the x-axis are $45^\circ, 105^\circ$ and $165^\circ$.

Hence proved.

With polar form,

On transforming the equation I get $$(\sin \theta - \cos \theta)(1 + 4\sin \theta \cos\theta) = 0$$

$$ \implies \tan\theta = 1, \sin 2\theta = -\frac12$$

$$\implies \theta = n\pi + \frac\pi4, {1 \over 2}\left(n\pi + (-1)^n\frac{7\pi}6\right) \tag{*}$$

But I don't know what to do with this now ? how can I show the lines are equally inclined with the help of (*) ?

Answer 1

Your calculation is correct, so just contrast them. Take $n=1,2,3$ $$\dfrac{7\pi}{12}-\dfrac{\pi}{4}=\dfrac{\pi}3$$ $$\dfrac{11\pi}{12}-\dfrac{7\pi}{12}=\dfrac{\pi}3$$. note: $0\leq\theta\leq\pi$.