\(\triangle \mathrm{OAB}\) is formed by the lines \(x^2-4 x y+y^2=0\) and the line \(A B\). The equation of line \(A B\) is \(2 x+3 y-1=0\). Then the equation of the median of the triangle drawn from the origin is

Let D be the midpoint of line AB .
\(\begin{array}{ll}
\therefore & \mathrm{A}=\left(x_1, y_1\right) \mathrm{B}=\left(x_2, y_2\right) \\
\therefore & \mathrm{D} \equiv\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)
\end{array}\)
Combined equation of side OA and OB is
\(x^2-4 x y+y^2=0\)
and equation of line AB is \(2 x+3 y-1=0\)
\(\therefore \quad\) Points A, B satisfy \(x^2-4 x y+y^2=0\) and
\(\begin{aligned}
& 2 x+3 y-1=0 \\
& \Rightarrow x=\frac{1-3 y}{2}
\end{aligned}\)
Substituting above value in \(x^2-4 x y+y^2=0\)
\(\begin{aligned}
\therefore \quad & \left(\frac{1-3 y}{2}\right)^2-4\left(\frac{1-3 y}{2}\right) y+y^2=0 \\
& \Rightarrow(1-3 y)^2-8 y(1-3 y)+4 y^2=0 \\
& 1-6 y+9 y^2-8 y+24 y^2+4 y^2=0 \\
\therefore \quad & 37 y^2-14 y+1=0 \\
& \text { Sum of roots }=\frac{-\mathrm{b}}{\mathrm{a}}=\frac{14}{37} \\
\therefore \quad & y_1+y_2=\frac{14}{37}
\end{aligned}\)
\(y \text {-coordinate of } \dot{\mathrm{D}}=\frac{y_1+y_2}{2}=\frac{7}{37}\)
Since point D lies on line AB
\(\therefore \quad\) Substituting \(y=\frac{7}{37}\) in \(2 x+3 y-1=0\)
\(\begin{aligned}
& \Rightarrow 2 x+3\left(\frac{7}{37}\right)-1=0 \\
& \Rightarrow 2 x+\frac{21}{37}-1=0 \\
& \Rightarrow x=\frac{8}{37} \\
\therefore & D \equiv\left(\frac{8}{37}, \frac{7}{37}\right)
\end{aligned}\)
Equation of median AD is
\(\begin{aligned}
& \frac{x-0}{0-\frac{8}{37}}=\frac{y-0}{0-\frac{7}{37}} \\
& \Rightarrow 7 x-8 y=0
\end{aligned}\)