smc

Cross-multiplying, we find that $(m+1)x^2-(bm+am+b-a)x+c(m-1)=0$. Because the roots of this quadratic are additive inverses, their sum is $0$. According to Vieta's Formulas, $\frac{bm+am+b-a}{m+1}=0$, or $m(a+b)=a-b$. Hence, $m=\boxed{\frac{a-b}{a+b}}$.