The formula of a midpoint of HX:
[tex]M\left(\dfrac{x_H+x_X}{2},\ \dfrac{y_H+y_X}{2}\right)[/tex]
We have the points
[tex]H\left(4\dfrac{1}{2},\ -4\deac{3}{4}\right),\ X\left(2\deac{1}{4},\ -2\dfrac{3}{4}\right)[/tex]
Substitute:
[tex]x_M=\dfrac{4\frac{1}{2}+2\frac{1}{4}}{2}=\dfrac{4\frac{1\cdot2}{2\cdot2}+2\frac{1}{4}}{2}=\dfrac{4\frac{2}{4}+2\frac{1}{4}}{2}=\dfrac{(4+2)+\frac{2+1}{4}}{2}\\\\=\dfrac{6\frac{3}{4}}{2}=6\dfrac{3}{4}:2=\dfrac{6\cdot4+3}{4}\cdot\dfrac{1}{2}=\dfrac{27}{4\cdot2}=\dfrac{27}{8}=3\dfrac{3}{8}\\\\y_M=\dfrac{-4\frac{3}{4}+\left(-2\frac{3}{4}\right)}{2}=\dfrac{-(4+2)-\frac{3+3}{4}}{2}=\dfrac{-6\frac{6}{4}}{2}=-6\dfrac{3}{2}:2\\\\=-\dfrac{6\cdot2+3}{2}\cdot\dfrac{1}{2}=-\dfrac{15}{4}=-3\dfrac{3}{4}[/tex]
Answer: [tex]M\left(3\deac{3}{8},\ -3\dfrac{3}{4}\right)[/tex]
