Distance formula calculator

Type two points, such as (1, 2) and (4, 8). Brackets and names are optional, and coordinates can be whole numbers, decimals, fractions like 3/4 or mixed numbers like 1 1/2. The distance appears as you type.

The answer is exact where it can be, in simplest radical form such as 3√5, with the decimal beside it. Below are the steps, the squared distance, the change in x and y, the midpoint, the slope and the equation of the line through both points.

The graph draws the segment from A to B with the right triangle under it: one leg across, one leg up, each labelled with its length. It scales to fit, so points like (−120, 40) and (300, 75) fit on it too. Type six numbers for two points in 3D; those get the working but no graph.

The distance between (x₁, y₁) and (x₂, y₂) is d = √((x₂ − x₁)² + (y₂ − y₁)²). Find how far apart the points are across and up, square both, add, and take the square root.

For (1, 2) and (4, 8): across is 4 − 1 = 3 and up is 8 − 2 = 6. Then d = √(3² + 6²) = √(9 + 36) = √45. The biggest square that divides 45 is 9, so √45 = √(9 × 5) = 3√5, about 6.708.

It doesn't matter which point comes first. Swapping them flips the signs of the differences, and squaring removes the signs again.

Draw a line across from A and a line up from B and they meet at a right angle. The segment AB is the long side of that right triangle, the hypotenuse. Its legs are the change in x and the change in y, so a² + b² = c² gives the distance. That's the triangle on the graph.

The US Common Core puts this in grade 8, as standard 8.G.B.8: apply the Pythagorean theorem to find the distance between two points in a coordinate system. In high school geometry, standard G-GPE.B.7 uses it to find the perimeters of shapes drawn on a grid.

For that reason the formula is sometimes called Pythagorean distance. Its proper name is Euclidean distance. Alexis Clairaut published the formula in 1731, almost a century after Descartes brought in coordinates in 1637.

In three dimensions the formula gains a z term: d = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²). It's Pythagoras twice, once across the floor and once up to the point. From (1, 2, 3) to (3, 5, 9) the differences are 2, 3 and 6, and √(4 + 9 + 36) = √49 = 7.