If ad neq 0 and two of the lines represented | vedclass

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(B) The given equation is $ax^3+3bx^2y+3cxy^2+dy^3=0$. Dividing by $x^3$ and letting $m = \frac{y}{x}$,we get the auxiliary equation: $dm^3+3cm^2+3bm+

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$a^2+3ac+3bd+d^2=0$

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(B) The given equation is $ax^3+3bx^2y+3cxy^2+dy^3=0$. Dividing by $x^3$ and letting $m = \frac{y}{x}$,we get the auxiliary equation: $dm^3+3cm^2+3bm+a=0$. Let the roots be $m_1, m_2, m_3$. The product of the roots is $m_1 m_2 m_3 = -\frac{a}{d}$. Since two lines are perpendicular,let $m_1 m_2 = -1$. Substituting this into the product,we get $(-1)m_3 = -\frac{a}{d}$,so $m_3 = \frac{a}{d}$. Since $m_3$ is a root of the auxiliary equation,it must satisfy it: $d(\frac{a}{d})^3 + 3c(\frac{a}{d})^2 + 3b(\frac{a}{d}) + a = 0$. Multiplying by $d^2$,we get $\frac{a^3}{d^2} + \frac{3ca^2}{d} + 3ba + ad = 0$. Multiplying by $d^2$ again (or simplifying the expression),we obtain $a^3+3a^2c+3abd+ad^2=0$. Dividing by $a$ (since $a 
eq 0$),we get $a^2+3ac+3bd+d^2=0$.

If $ad \neq 0$ and two of the lines represented by $ax^3+3bx^2y+3cxy^2+dy^3=0$ are perpendicular,then

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