I.      If  two of the lines represented by $a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3=0(a\ne 0)$ make complementary angles with x-axis anticlock         wisesense then slope of third line is a/d

II.           If  the  slope of one of the line represented by $a{x}^2+2hxy+b{y}^2=0$ is ‘n’ times the slope of  another then          $\frac{{(1+n)}^2}{n}=\frac{h^2}{ab}$

 Which of the above statement is correct :

$\begin{array}{l}GivenEquationa{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3=0\\ Lety=mxbetheoneoftheline\\ \Rightarrow a{x}^3+b{x}^2\left(mx\right)+cx\left(m^2{x}^2\right)+d\left(m^3{x}^3\right)=0\\ \Rightarrow x^3(a+bm+c{m}^2+d{m}^3)=0\\ \therefore d{m}^3+c{m}^2+bm+a=0\\ Let{m}_1,m_2,m_{3}areslopesthen\\ m_1{m}_2{m}_3=\frac{-a}{b}\\ Butgiven{m}_1{m}_2=-1\\ \Rightarrow -1\left(m_3\right)=\frac{-a}{d}\\ \Rightarrow m_3=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$d$}\right.\\ True\end{array}$

$\begin{array}{l}IIGivenEquationa{x}^2+2hxy+b{y}^2=0\\ letm,nmareslopes\\ sumofslopes\Rightarrow m+nm=\frac{-2h}{b}\\ \Rightarrow m(1+n)=\frac{-2h}{b}\\ \Rightarrow m=\frac{-2h}{b(1+n)}\\ \mathrm{Pr}oductofslopes=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\\ \Rightarrow m\left(nm\right)=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\\ \Rightarrow n\left(m^2\right)=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\\ \Rightarrow n\left(\frac{4h^2}{b^2{(1+n)}^2}\right)=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\\ \Rightarrow \frac{4h^2}{ab}=\frac{{(n+1)}^2}{n}\\ True\end{array}$