1. **State the problem:** We need to find the midpoints of the line segments AB, EF, and CE, and the distances between points AC, BC, AF, and DF given the points: A(1,2), B(3,4), C(5,2), D(3,-2), E(4,4), F(1,-3). 2. **Midpoints:** The midpoint $M$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$M=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$ ① Midpoint of AB: $$=\left(\frac{1+3}{2}, \frac{2+4}{2}\right) = (2,3)$$ ② Midpoint of EF: $$=\left(\frac{1+4}{2}, \frac{-3+4}{2}\right) = \left(\frac{5}{2}, \frac{1}{2}\right) = (2.5, 0.5)$$ ③ Midpoint of CE: $$=\left(\frac{5+4}{2}, \frac{2+4}{2}\right) = \left(\frac{9}{2}, 3\right) = (4.5, 3)$$ 3. **Distances:** The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ ④ Distance AC: $$= \sqrt{(5-1)^2 + (2-2)^2} = \sqrt{4^2 + 0} = \sqrt{16} = 4$$ ⑤ Distance BC: $$= \sqrt{(5-3)^2 + (2-4)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.828$$ ⑥ Distance AF: $$= \sqrt{(1-1)^2 + (-3-2)^2} = \sqrt{0 + (-5)^2} = \sqrt{25} = 5$$ ⑦ Distance DF: $$= \sqrt{(1-3)^2 + (-3+2)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236$$ **Final answers:** - Midpoints: AB = (2,3), EF = (2.5,0.5), CE = (4.5,3) - Distances: AC = 4, BC = $2\sqrt{2}$, AF = 5, DF = $\sqrt{5}$