Two of the lines represented by the equation $a y^4+b x y^3+c x^2 y^2+d x^3 y+e x^4=0$ will be perpendicular…
Two of the lines represented by the equation $a y^4+b x y^3+c x^2 y^2+d x^3 y+e x^4=0$ will be perpendicular, then
$(b+d)(a d+b e)+(e-a)^2(a+c+e)=0$
$(b+d)(a d+b e)+(e+a)^2(a+c+e)=0$
$(b-d)(a d-b e)+(e-a)^2(a+c+e)=0$
$(b-d)(a d-b e)+(e+a)^2(a+c+e)=0$
Solution
Let $a y^4+b x y^3+c x^2 y^2+d x^3 y+e x^4$
$=\left(a x^2+p x y-a y^2\right)\left(x^2+q x y+y^2\right)$
Comparing the coefficients of similar terms,
b = aq − p, c = − pq, d = aq + p and e = − a
b + d = 2aq and e − a = − 2a
ad + be = 2ap and a + c + e = − pq
$\begin{aligned} & \because(b+d)(a d+b e)=(2 a q) \times 2 a p=4 a^2 p q \\ & \text { Also, }-(e-a)^2(a+c+e)=-(-2 a)^2(-p q)=4 a^2 p q \\ & \therefore(b+d)(a d+b e)=-(e-a)^2(a+c+e) \\ & \Rightarrow(b+d)(a d+b e)+(e-a)^2(a+c+e)=0\end{aligned}$