Answer
$\dfrac{-4}{5}$
Work Step by Step
Since, we have $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ ...(1)
Now, need to plug the derivatives in equation (1).
we have $\dfrac{2x+y}{2y+x} \times (-1)=\dfrac{-2x-y}{2y+x} $
Now, plug the point $(1,2)$
we get $\dfrac{-2x-y}{2y+x} =\dfrac{-2(1)-2}{2(2)+1} =\dfrac{-4}{5}$
