[tex]\begin{gathered} \text{let's remember how to calculate the midpoint:} \\ p=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\text{ } \\ so\text{ in this case we have} \\ p=(\frac{97+-12}{2},\frac{8+202}{2})=(\frac{85}{2},\frac{210}{2})=(42.5,105) \\ So\text{ the answer is no, M is not the midpoint of AB} \end{gathered}[/tex]
