If $2x^2-5xy+2y^2=0$ represents two sides of a triangle and centriod is (2, 3) then equation of third side is

Let  $f=2x^2-5xy+2y^2=0andG=\left(2,3\right)$

Mid point of Third side

$\begin{array}{l}=\frac{3}{2}G\\ =\frac{3}{2}\left(2,3\right)=\left(3,\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\\ \frac{\partial f}{\partial x}=4x-5y\frac{\partial f}{\partial y}=4y-5x\end{array}$

Slope of Third side

$\begin{array}{l}=-\left(\frac{\raisebox{1ex}{$\partial f$}\!\left/ \!\raisebox{-1ex}{$\partial y$}\right.}{\raisebox{1ex}{$\partial f$}\!\left/ \!\raisebox{-1ex}{$\partial y$}\right.}\right)G\\ =-\left(\frac{4x-5y}{4y-5x}\right)G\\ =-\left(\frac{4\left(3\right)-5\left(\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{4\left(\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)-5\left(3\right)}\right)\\ =-\left(\frac{12-\raisebox{1ex}{$45$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{18-15}\right)\\ =-\left(\frac{24-45}{2\left(3\right)}\right)=\frac{21}{6}=\frac{7}{2}\end{array}$

Equation of third side is

$\begin{array}{l}y-\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{7}{2}\left(x-3\right)\\ \Rightarrow 2y-9=7\left(x-3\right)\\ \Rightarrow 2y-9=7x-21\\ \Rightarrow 7x-2y-12=0\end{array}$