The equation of the pair of perpendicular lines passing through origin and forming an isosceles triangle with the line 2x+ 3y=6 is

$\begin{array}{l}Lety=mxbetheoneofline\\ m=\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\Rightarrow m_1=\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\\ andgivenline2x+3y-6=0\\ Slope=\frac{-2}{3}\Rightarrow m_2=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ \mathrm{Tan}\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\\ \mathrm{Tan}{45}^0=\left|\frac{\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{1+\left(\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)\left(\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}\right|\\ =\left|\frac{3y+2x}{3x-2y}\right|\\ \Rightarrow \left|3x-2y\right|=\left|2x+3y\right|\\ S.B.S\\ \Rightarrow 9x^2+4y^2-13xy=4x^2+9y^2+13xy\\ \Rightarrow 5x^2-24xy-5y^2=0\end{array}$