Distance & Midpoint Formulas

Two formulas cover most coordinate geometry.

Distance: d = √((x₂ − x₁)² + (y₂ − y₁)²)

That is just the Pythagorean theorem. The horizontal gap and the vertical gap are the legs of a right triangle, and the distance is the hypotenuse.

Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2)

Average the x's, average the y's. The answer is a point, so it needs both coordinates.

Because both differences get squared, the order of subtraction does not matter for distance. It very much does for slope.

A common variant gives you the midpoint and one endpoint and asks for the other. Do not average — work backwards: the midpoint is as far from the known endpoint as the unknown one is, so double the midpoint and subtract.

The distance formula is the Pythagorean theorem applied to the legs |x₂ − x₁| and |y₂ − y₁|, so d = √((x₂−x₁)² + (y₂−y₁)²). Squaring removes any sign dependence, making subtraction order irrelevant.

The midpoint is the componentwise arithmetic mean, ((x₁+x₂)/2, (y₁+y₂)/2), the t = ½ case of the section formula (x₁ + t(x₂−x₁), y₁ + t(y₂−y₁)) which locates any point dividing a segment in a given ratio.

For the missing-endpoint problem, solving (x₁ + x₂)/2 = m_x gives x₂ = 2m_x − x₁.

Both formulas extend to three dimensions by adding a z term, and the distance formula generalises to any dimension as the Euclidean norm.

Together with slope, these are the three tools of coordinate proof: distance settles congruence, midpoint settles bisection, and slope settles parallelism and perpendicularity. Choosing which to compute — rather than computing all of them — is what makes a coordinate proof efficient.