# Analytic geometry straight line problem

Prove that two straight lines represented by the equation $x^3+y^3+bx^2y+cxy^2=0$ will be at right angles if $b+c=-2$

I didn't know that even straight lines like planes can be represented by a combined equation. can someone please explain how this happens and how to find the individual lines so that the angle between them may be determined.

Thanks

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Well, this equation doesn't always represent two lines. It could also be one line or three lines. For instance, like this‌​. – Dan Shved Apr 30 '13 at 11:47
@DanShved, I think that would imply the equation is equivalent to $(x+ry+s)^2(x+uy+v)=0$. (If his question is correct.) – Easy Apr 30 '13 at 11:49
@Easy Yep. But still, should we assume that there are indeed exactly two lines? Or does the question also asks us to prove this (when $b+c=-2$)? To sum up, I don't like the phrasing of the question ) – Dan Shved Apr 30 '13 at 11:51
How do we find combined equations of two or more straight lines ? like for plane we just do $\phi(x,y,z)*\theta(x,y,z)=0$ – Aman Mittal Apr 30 '13 at 11:51
I thought the meaning of the question could be that $\,aw^2+bz^2=0\implies a=0=b\,$ , and from here the two lines...? – DonAntonio Apr 30 '13 at 11:58

Note that the equation is homogeneous, so we let $x^3+y^3+bx^2y+cxy^2=(x+hy)^2(x+ky)$. Since we want two lines $x+hy=0,x+ky=0$ perpendicular, so $$(-\frac{1}{h})(-\frac{1}{k})=-1\Rightarrow hk=-1.$$Equating the corresponding coefficients gives $$h^2k=1,~2h+k=b,~h^2+2hk=c\Rightarrow h=-1,k=1\Rightarrow b=c=-1$$Thus, we have $b+c=-2$.

The converse: Assuming $b+c=-2$ we have $h^2+2hk+2h+k=-2$, that is $$(h^2+2h+1)+(1+2hk+k)=0\Leftrightarrow(h+1)^2+(h^2k+2hk+k)=0\Leftrightarrow(h+1)^2(k+1)=0$$So $h=-1,~k=-1$. But $h^2k=1$ forces that $h=-1,k=1$.

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You have shown that if the lines are orthogonal, then $b+c=-2$. The question asked about the opposite direction. – MvG Apr 30 '13 at 13:30
@MvG, in fact we can obtain it as a necessary and sufficient condition. – Easy Apr 30 '13 at 14:02
Thanks for this solution. I have silly question though. A first degree equation in x,y,z represents a plane. And a pair of plane represents a line through their intersection. so how did you take equation of two lines as $(x+hy)$ and $(a+ky)$ ? Should they rather not be equation of two planes? am i missing out on something ? – Aman Mittal Apr 30 '13 at 14:16
@AmanMittal, the equation only involves $x,y$, so the graph should be in $xy$-plane, in which case the general line equation is $Ax+By=C$. – Easy Apr 30 '13 at 14:20
Ah !! Thats the point. Thanks a lot, i was working on 3D too much lately and missed out this simple point. Thanks !! – Aman Mittal Apr 30 '13 at 14:21

If by "lines represented by..." is meant that

$$0=x^3+y^3+bx^2y+cxy^2=x^2\color{red}{(x+by)}+y^2\color{red}{(y+cx)}\implies$$

the lines $by+x=0\;,\;\;y+cx=0\,$ are perpendicular to each other iff the product of their slopes is minus one, i.e. iff

$$-\frac1b\cdot (-c)=-1\iff -b=c\iff b+c=0$$

so either I'm misunderstanding the meaning of "representing the lines..." or else there's a mistake in the claim.

BTW, at the beginning,

$$b=0\implies x=0\;,\;\;\text{and then it should be}\;y=0\iff c=0\;\text{for the lines to be perpendicular}$$

as both the lines have zero free coefficient

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Thanks for the reply. Can you please explain how can we pick the red portion of the equation (written by you) to be two lines ? aren;t they equations of planes anyway – Aman Mittal Apr 30 '13 at 12:00
@AmanMittal, I explained my mind in the comments below the question: my thought is that perhaps what is meant here is to use the relation for non-negative real numbers $\,aw^2+bz^2=0\implies a=b=0\,$ , with $\,a,b\,$ being, in our case, the expressions in red in my answer. I'm not completely sure whether this is what is meant in your question, though... – DonAntonio Apr 30 '13 at 12:03
Yet inputting $\,x^3+y^3+3x^y-5xy^2=0\,$ in WA gives indeed three lines and not only two, just as one of the users already mentioned... – DonAntonio Apr 30 '13 at 12:07
You should be multiplying the expressions for the individual lines, not adding them as you seem to be doing in your first equation. – MvG Apr 30 '13 at 13:48
@MvG, that's why I explained what I tried to do. Since the usual did not give two but three lines (at least in several cases), Mine is an alternative approach. I still am not sure what the OP really meant... – DonAntonio Apr 30 '13 at 14:22

Suppose you have two lines through the origin, one of slope $\alpha$ and the other of slope $\beta$:

\begin{align*} l_1:\quad y &= \alpha x & \alpha x - y &= 0 \\ l_2:\quad y &= \beta x & \beta x - y &= 0 \end{align*}

Then you can multiply (powers of) these euqations to obtain a combined equation, which will be zero if and only if one of its factor polynomials is zero:

\begin{align*} l_1^2 \cdot l_2&: & \alpha^{2} \beta x^{3} - (\alpha^{2} + 2 \, \alpha \beta) x^{2} y + (\beta + 2 \, \alpha) x y^{2} - y^{3} &= 0\\ l_1 \cdot l_2^2&: & \alpha \beta^{2} x^{3} - (\beta^{2} + 2 \, \alpha \beta) x^{2} y + (\alpha + 2 \, \beta) x y^{2} - y^{3} &= 0 \end{align*}

But these equations don't have the required form. It is enough to concentrate on the first equation, since $l_1$ and $l_2$ are structurally equivalent, so you might as well swap them. To turn $l_1^2\cdot l_2$ into the required form, you should multiply it with $-1$ in order to get $+y^3$ instead of $-y^3$. Then you get a monomial of $+x^3$ exactly if $\alpha^2\beta=-1$. So we now want

\begin{align*} \alpha^2\beta &= -1 & \\ \alpha^2 + 2\alpha\beta &= b & \\ \beta + 2\alpha &= -c \\ b + c &= -2 \end{align*}

Typing this in e.g. Wolfram Alpha you will get one real and two complex solutions:

\begin{align*} b &= -1 & b &= -1-2i & b &= -1+2i \\ c &= -1 & c &= -1+2i & c &= -1-2i \\ \alpha &= 1 & \alpha &= -i & \alpha &= i \\ \beta &= -1 & \beta &= 1 & \beta &= 1 \\ \alpha\beta &= -1 & \alpha\beta &= -i & \alpha\beta &= i \end{align*}

The real solution satisifes your requirement: the slopes multiply to $-1$, so the lines are orthogonal. Moreover, there is only a single pair of lines with that property. The complex solutions don't satisfy your requirement, but you'll probably want to exclude these from your consideration in any case.

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