Distance and Midpoint Between Two Points

Calculates the distance between two points (x₁, y₁) and (x₂, y₂) as √((x₂−x₁)² + (y₂−y₁)²).

Explanation

This finds the straight-line distance between two points on the coordinate plane, along with their midpoint. The distance formula is the Pythagorean theorem in disguise: picture a right triangle whose legs are the horizontal gap and the vertical gap, and the distance is its hypotenuse.

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The midpoint is nothing more than the average of the two coordinates.

M = \left( \dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2} \right)

$(x_1, y_1)$ , $(x_2, y_2)$ — the two points
$d$ — the distance between them
$M$ — the midpoint, cutting the segment exactly in half

Example

For $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (3, 4)$ :

d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

The distance is 5, and the midpoint is $\left( \dfrac{0 + 3}{2},\ \dfrac{0 + 4}{2} \right) = (1.5,\ 2)$ .

Notes

The differences get squared, so $x_2 - x_1$ and $x_1 - x_2$ give the same answer. The order of the points does not matter.
Negative coordinates behave no differently. From $(-2, 1)$ to $(1, 5)$ the gaps are 3 and 4, so the distance is 5.
A distance is never negative, and it is zero only when the two points coincide.
In three dimensions, add the squared difference in $z$ under the same square root. Nothing else changes.