The slopes of three sides of a triangle ABC are -1, -2, 3 respectively. If one orthocenter of triangle ABC is origin, then the locus of the centroid is  $y=\frac{a}{b}x$  where a, b are respectively prime then $b–a$ is equal to 

Let the slopes of BC, CA, AB be -1, -2, 3 respectively

$=slopeofAH=1\Rightarrow A=(x_1,x_1)$

Slope of  $BH=\frac{1}{2}\Rightarrow B=(x_2,\frac{1}{2}{x}_2)$
Slope of  $CH=-\frac{1}{3}\Rightarrow C=(x_3,{\frac{x_3}{3}}_2)$
Slope of  $BC=\frac{\frac{x_2}{2}+\frac{x_3}{3}}{x_2-x_2}=\frac{3x_2+2x_3}{6(x_2-x_3)}$
$$\begin{gathered} =-1\Rightarrow 9x^2=4x_3 \\ \therefore slopeofCA=\frac{x_1+\frac{x_3}{3}}{x_1-x_3}=-2 \\ =\frac{3x_1+x_3}{3(x_1-x_3)}\Rightarrow 9x_1=5x_3 \\ G=(h,k)=(\frac{x_1+x_2+x_3}{3},\frac{x_1+\frac{x_2}{2}-\frac{x_3}{3}}{3}) \\ 3h=x_3+\frac{5x_3}{9}+\frac{4x^3}{9}=2x_3 \dots \dots \dots \dots \dots \left(1\right) \\ 3k=\frac{5x_3}{9}+\frac{2x_3}{9}-\frac{x_3}{3}=\frac{4x_3}{9} \dots \dots \dots \dots \dots .\left(2\right) \end{gathered}$$

From equations (1) and (2) we get 

$\frac{k}{h}-\frac{2}{9}\Rightarrow y=\frac{2}{9}x;b-a=9-2=7$