0

The question was to prove that the equation $y^3-x^3+3xy(y-x)=0$ represents three lines equally inclined to one another.What i did was i factored out $(y-x)$ and the equation became $$(y-x)(y^2+4xy+x^2)$$. Now, since the other part is a homogeneous equation of second degree, so i can safely say that it represents two straight lines. Now i guessed that if the lines are equally inclined then the bisector of the lines given by $y^2+4xy+x^2$ will be the line $y-x=0$. So, going forward to prove it i tried to find the bisectors by using the formula $(x^2-y^2)/(a-b)=xy/h$, where a and b are the coefficients of $x^2$ and $y^2$ and h is the coeffient of the term $2xy$. The equations of the bisectors were $y-x$ and $y+x$. which matched the factored out term in the original equation.So, my question is that , is this enough to complete the proof?

Ayan Shah
  • 491
  • Maybe easier to note that the homogeneous quadratic is symmetric in $x,y$, so the two lines must be symmetric with respect to $y=x$. – dxiv Oct 17 '16 at 04:35

1 Answers1

1

1) The formula used should be $hx^2 –(a – b)xy – h^2 = 0$ instead. This is because one has to avoid the $a – b$ term being the zero denominator.

2) It turns out one of the angle bisectors is $L_1 : y = x$. This means both $L_2$ & $L_3$ are equally inclined to $L_1$.

3) We still need to prove that $L_2$ or $L_3$ are also equal inclined to $L_1$.

This can be done by:-

(a) Finding the angle between $L_2$ and $L_3$ via the formula $\tan \theta = \dfrac {2\sqrt {h^2 - ab}}{a + b}$. (It turns out that $\theta = 60^0$); and

(b) Further factorizing $y^2+4xy+x^2$ to $[(y + 2x) + \sqrt 3 x][ (y + 2x) - \sqrt 3x]$. Then, find the angle between $L_1$ and $L_2$.

Mick
  • 17,141
  • I did as you said and the angles between L1 and L2 and also between L3 and L1 turned out to be 60 degree. But now i am confused with the geometric representation of this, as adding all the angles gives me 180 degrees whereas since i am working with 2 dimensional plane i should get 360 degress. – Ayan Shah Oct 17 '16 at 06:06
  • Angle b/w (L1 & L2) = angle b/w (L2 & L3) and angle b/w (L3 & L1) $= 60^0$ . They are all on a 2D plane. If vertically opposite angles are also counted, then there are $360^0$. – Mick Oct 17 '16 at 06:13
  • But L1 is in between L2 and L3 , so how can an angle between L1 and L3 be equal to angle between L2 and L3 – Ayan Shah Oct 17 '16 at 06:16
  • because they are all $60^0$. 2) a line has two ends.
    • – Mick Oct 17 '16 at 06:23
  • OOOOO....now i get it, thank you for opening up a new perspective in me. Highly appreciated – Ayan Shah Oct 17 '16 at 06:25