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
 A: 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 $h=-1,k=1$. 
A: 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 
A: 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.
