Equation of three lines passing through the origin If $dx^3+7x^2y+bxy^2+y^3=0$ represent three straight lines of which two of them makes complementary angles with x-axis then the value of $|d^2-bd|$=_______
My approach is as follows
From the above equation, presume that line passes through origin
The equation of lines are $y=m_1x$, $y=m_2x$ & $ y=m_3x$
For complementary angle the angle between them is $180^o$
$(y-m_1x)(y-m_2x)(y-m_3x)=0$
$(y^2-(m_1+m_2)xy+m_1m_2x^2)(y-m_3x)=0$
$(y^3+(m_1m_2+m_2m_3+m_3m_1)x^2y-(m_1+m_2+m_3)xy^2+m_1m_2m_3x^3)=0$
How do I proceed from here. For supplementary angle $m_1m_2=-1$  then what is the condition for complementary angles?

$m_1m_2=1$. Let $m_1 + m_2 = S$.
Comparing coefficients,
$$y^3 - xy^2\sum m_1  + x^2y \sum m_1m_2 \color{red}{-x^3}m_1m_2m_3  = 0 =$$$$ y^3 +  bxy^2 + 7x^2y + dx^3$$
we get
$$ d=-m_3$$
$$-b = S+m_3$$
$$ 7 = 1 + m_3S$$
so that $$|d(d-b)| = |-m_3(S)| $$
$$ = 6 $$
