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

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    What does this mean: "To be at right angles of $a+b=-2$? Perhaps: perpendicular to the line $ax+by=-2$? – 2017-02-12
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    @zoli I imagine it's supposed to be "... *if* $b+c=-2$." – 2017-02-12
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    OP, since there is some contention, do you mean **two pairs of lines** will be at right angles, or **two lines** will be at right angles? – 2017-02-13
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    The question is from an academic exam ; it's given pairs only – 2017-02-15

2 Answers 2

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Since a homogeneous polynomial in two variables factors into linear factors, then we can write $$ f = x^3+byx^2+cxy^2+y^3 = (y - m_1 x) (y-m_2x)(y-m_3x) \, . $$ Multiplying out and equating coefficients yields $3$ equations: \begin{align} -m_1 m_2 m_3 &= 1 \tag{1}\\ m_1 m_2 + m_1 m_3 + m_2 m_3 &= b \tag{2}\\ -m_1 - m_2 - m_3 &= c \tag{3} \end{align} and we also have the equation \begin{align} b + c &= -2 \, . \tag{4} \end{align} From $(4)$ we have $b = -c - 2$, so combining $(2)$ and $(3)$ yields $$ m_1 m_2 + m_1 m_3 + m_2 m_3 = b = - c - 2 = m_1 + m_2 + m_3 - 2 $$ i.e., $$ m_1 m_2 + m_1 m_3 + m_2 m_3 - m_1 - m_2 - m_3 + 2 = 0 \, . $$ Using $(1)$ to solve for $m_1$ we have $m_1 = -\frac{1}{m_2 m_3}$, and substituting it into the expression above, we find \begin{align*} 0 &= -\frac{1}{m_2 m_3} m_2 - \frac{1}{m_2 m_3} m_3 + m_2 m_3 + \frac{1}{m_2 m_3} - m_2 - m_3 + 2\\ &= -\frac{1}{m_3} - \frac{1}{m_2} + m_2 m_3 + \frac{1}{m_2 m_3} - m_2 - m_3 + 2 \, . \end{align*} Multiplying through by $m_2 m_3$ to clear denominators yields \begin{align*} 0 &= m_2^2 m_3^2 - m_2^2 m_3 - m_2 m_3^2 + 2 m_2 m_3 - m_2 - m_3 + 1 = (m_{3} - 1) (m_{2} - 1) (m_{2} m_{3} + 1) \, . \end{align*} Thus $m_2 = 1$ or $m_3 = 1$ or $m_2 m_3 = -1$. In the third case, we have $m_2 = -\frac{1}{m_3}$, so the slopes of the lines are negative reciprocals, hence the lines are perpendicular. If $m_2 = 1$, then from $(1)$ we have $-m_1 m_3 = 1$, so $m_3 = -\frac{1}{m_1}$, and the case $m_3 = 1$ can be resolved in the same way.

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This equation represents three lines through the origin. Two of them must be identical and perpendicular to the third.

The slope equation must have a double root which is the opposite of the inverse of the third.

$$1+bt+ct^2+t^3=(t-m)^2\left(t+\frac1m\right).$$

By identification

$$m=1,-2+m^2=b,-2m+\frac1m=c,$$

$$b=-1,\\c=-1.$$

The straight lines are

$$\begin{cases}y=x,\\y=x,\\y=-x.\end{cases}$$

corresponding to

$$x^3-x^2y-xy^2+y^3=0.$$

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    It is not true that two of the lines are always identical. The problem statement just says that two of the three lines are perpendicular. [Here is an example where two of the lines are not identical.](https://www.wolframalpha.com/input/?i=x%5E3%2B(0)yx%5E2%2B(-2)xy%5E2%2By%5E3%3D0): take $b=0$ and $c = -2$. – 2017-02-12
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    @SpamIAm: the problem statement says **two pairs**. Following this requirement, your answer is wrong. – 2017-02-13
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    The problem is misstated and the only reasonable interpretation is that two of the three lines are perpendicular. Your answer cannot be right: I have shown with an example that even if $b+c = -2$, it is not necessarily the case that two of the lines coincide. I also find your downvote quite discourteous. – 2017-02-13
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    @SpamIAm: I am answering the question that is asked and not extrapolating. "the only reasonable interpretation" is your saying. If you doubt about the problem statement, you should tell the OP. – 2017-02-13
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    The problem as stated is *false*. The problem statement says: if $b + c = -2$, then two pairs of the straight lines will be at right angles. You cannot prove this, because it is *false*, as shown by the counterexample I stated above. Do you not agree that this is a counterexample? And I left a comment asking for clarification from the OP before you replied. – 2017-02-13
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    @SpamIAm: I showed a solution with $b=c=-1$, which is such that $b+c=-2$, if I am right. You asked for clarification half an hour *after* my comment. – 2017-02-13
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    I meant your most recent reply. I guess I need to make the quantifiers in the problem statement clearer. The problem, as stated, is *for any* $b$ and $c$, if $b + c = -2$, then two pairs of the lines are perpendicular. You have shown that the conclusion is true for *one particular choice* of $b$ and $c$, not general $b$ and $c$. – 2017-02-13