Answer
$2$
Work Step by Step
Consider $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$
Plug the derivatives in equation $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$
This implies that $\dfrac{(y-3)}{(x+2y)} (-1)=\dfrac{3-y}{(x+2y)} $
For point $(-1,1)$, we have
$\dfrac{3-y}{x+2y} =\dfrac{3-1}{-1+2(1)}=2$
