Two jobs on the grid today, both on two points: $(x_1, y_1)$ for the first landmark and $(x_2, y_2)$ for the second. The midpoint is pure fairness: average the x's, average the y's: $x_m = \dfrac{x_1 + x_2}{2}$, and the same again for y. Meet in the middle, coordinate by coordinate. (The classic slip is halving the GAP instead of averaging the ENDS — that gives how far the middle is, not where it is.)

The distance between two points is old wisdom in new clothes: the run and the rise make the two legs of a right triangle, and Pythagoras does the rest — $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$, d equals the square root of run-squared plus rise-squared — where $\Delta x = x_2 - x_1$ is the run and $\Delta y = y_2 - y_1$ is the rise ($\Delta$ is delta, and it only ever means “the change in”). Legs 3 and 4? The span is 5. The 3-4-5 family and its cousins (6-8-10, 5-12-13) will follow you all the way to the trigonometry chapter — learn their faces now and they'll wave back later.