[tex]\text{The formula of a midpoint:}\\\\\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)\\\\\text{We have}\ R(-2b,\ 2c)\ \text{and}\ T(-2a,\ 0).\ \text{Substitute:}\\\\x=\dfrac{-2b+(-2a)}{2}=\dfrac{-2b}{2}+\dfrac{-2a}{2}=-b-a\\\\y=\dfrac{2c+0}{2}=\dfrac{2c}{2}=c\\\\Answer:\ \boxed{b.\ (-a-b,\ c)}[/tex]
