The slopes of sides of a triangle are -1, -2, 3. If the orthocenter of the triangle is the origin O; then the locus of its centroide is y/x =

$\begin{array}{l}LetslopeofBC=-1\\ slopeofAC=-2\\ slopeofAB=3\\ OA\perp BC\Rightarrow slopeofOA=1\Rightarrow A(\alpha ,\alpha )\\ OB\perp AC\Rightarrow slopeofOB=\frac{1}{2}\Rightarrow B(2\beta ,\beta )\\ \overline{OC}\perp AB\Rightarrow slopeof\overline{BC}=\frac{-1}{3}\\ \therefore C(-3\gamma ,\gamma )\\ \therefore slopeofBC=-1\\ \Rightarrow \frac{\beta -\gamma }{2\beta +3\gamma }=-1\\ \Rightarrow \beta -\gamma =-2\beta -3\gamma \end{array}$

$\begin{array}{l}\Rightarrow 3\beta =-2\gamma \\ \Rightarrow \gamma =\frac{-3\beta }{2}\\ slopeofAB=3\\ \therefore \frac{\beta -\gamma }{2\beta -\alpha }=3\\ \Rightarrow \beta -\alpha =6\beta -3\alpha \\ \Rightarrow 2\alpha =5\beta \\ \Rightarrow \alpha =\frac{5\beta }{2}\\ G=(\frac{\alpha +2\beta -3\gamma }{3},\frac{\alpha +\beta +\gamma }{3})\\ =(\frac{\frac{5\beta }{2}+2\beta +\frac{9\beta }{2}}{3},\frac{\frac{5\beta }{2}+\beta -\frac{3\beta }{2}}{3})\\ =(\frac{9\beta }{3},\frac{2\beta }{3})=(x,y)\\ \therefore \frac{y}{x}=\frac{2}{9}\end{array}$