In Distance Between Two Points you found the distance between two points by drawing a right triangle under the segment and using the Pythagorean theorem. In Points, Lines and Planes you met $AB$ as the length of a segment. Here we turn that picture into a formula, and add a second one for the point exactly halfway between, the midpoint.
Why it works
Distance. From $A(x_1, y_1)$ to $B(x_2, y_2)$, walk across $x_2 - x_1$ and up $y_2 - y_1$. Those are the legs of a right triangle and $\overline{AB}$ is the hypotenuse, so $AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$. Squaring also takes care of direction: a step of $-4$ and a step of $4$ both give $16$.
Midpoint. Halfway along the segment, you have gone half of the way across and half of the way up. Half of the way from $x_1$ to $x_2$ lands on their average, $\frac{x_1 + x_2}{2}$; the same is true for $y$.
Midpoint. Halfway along the segment, you have gone half of the way across and half of the way up. Half of the way from $x_1$ to $x_2$ lands on their average, $\frac{x_1 + x_2}{2}$; the same is true for $y$.
Key idea
$$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ The distance is the Pythagorean theorem; the midpoint is the average of the endpoints.
Worked example: An exact distance
Find the distance between $A(-2, 3)$ and $B(4, -1)$.
- Changes: $x_2 - x_1 = 4 - (-2) = 6$ and $y_2 - y_1 = -1 - 3 = -4$.
- Square and add: $6^2 + (-4)^2 = 36 + 16 = 52$.
- $AB = \sqrt{52} = \sqrt{4 \cdot 13} = 2\sqrt{13} \approx 7.2$.
Answer: $AB = 2\sqrt{13}$
Worked example: A midpoint
Find the midpoint of the segment from $A(-3, 5)$ to $B(7, -2)$.
- Average the $x$-coordinates: $\frac{-3 + 7}{2} = \frac{4}{2} = 2$.
- Average the $y$-coordinates: $\frac{5 + (-2)}{2} = \frac{3}{2}$.
- $M = \left(2, \frac{3}{2}\right)$. Check: it lies between the endpoints in both directions.
Answer: $\left(2, \frac{3}{2}\right)$
Worked example: Working backward to an endpoint
$M(1, -2)$ is the midpoint of $\overline{AB}$ and $A = (-4, 3)$. Find $B$.
- From $A$ to $M$: $x$ goes up by $1 - (-4) = 5$ and $y$ goes down by $3 - (-2) = 5$.
- $M$ is halfway, so repeat the same step from $M$: $B = (1 + 5, -2 - 5) = (6, -7)$.
- Check: $\frac{-4 + 6}{2} = 1$ and $\frac{3 + (-7)}{2} = -2$.
Answer: $B = (6, -7)$
Common mistake
Two slips to watch for. (1) Subtracting for a midpoint: $\frac{x_2 - x_1}{2}$ is half of the change, not a location. Check: the midpoint must lie between the two endpoints. (2) Dropping the sign before squaring: $(-4)^2 = 16$, not $-16$. Check: the distance is always at least as big as the larger of the two changes.