Distance Calculator - Distance Between Two Points (2D and 3D)

A distance calculator finds how far apart two points are in the Euclidean plane or in space. In two dimensions the formula is the Pythagorean theorem on the run and the rise: d = √[(x2−x1)² + (y2−y1)²]. In three dimensions you add the third square: + (z2−z1)² under the same root. The result is a length, never negative.

The default pair is (0, 0) and (3, 4). Then d = √(9+16) = 5, the classic 3-4-5 hypotenuse. The midpoint is (1.5, 2). The slope is 4/3. Coordinates may be negative; (−3, 0) to (0, 4) is still 5. Coincident points give distance 0, which is a valid measurement, not an error.

This is not geodesic distance on a map (great-circle miles), not Manhattan |Δx|+|Δy|, and not driving directions. If you only needed the slope of the segment, the slope calculator goes further (angle, intercept, equation). If you have two legs and want the hypotenuse without coordinates, use the Pythagorean theorem calculator. A full right-triangle solve with angles belongs on the right triangle calculator.

Using this 2D 3D point distance tool takes four steps:

1. Choose 2D or 3D. Slope is reported only in 2D. Midpoint always uses the coordinates you entered.
2. Type both points. Empty fields wait. Non- numeric text errors. Zero is allowed.
3. Read d, Δx Δy (Δz), midpoint, and slope (or “vertical” when x1 = x2 in the plane).
4. Copy or clear. Points stay in localStorage for up to 30 days.

Note: there is no unit dropdown. If x is in metres, d is in metres. GPS latitude/longitude is not Euclidean without a projection; do not paste degrees here and expect kilometres.

Worked example - (0, 0) to (3, 4):

• Δx = 3, Δy = 4
• d = √(9 + 16) = √25 = 5
• Midpoint ((0+3)/2, (0+4)/2) = (1.5, 2)
• Slope m = 4/3

3² + 4² = 5² is the smallest primitive Pythagorean triple besides order. Any similar segment - (1, 1) to (4, 5), still Δ = (3, 4) - has the same length 5. Multiples 6-8-10 and 9-12-15 are the same family scaled.

Worked example - vertical segment (2, 1) to (2, 7):

Δx = 0, Δy = 6, d = 6. Slope is undefined (vertical). Distance is still well-defined. A horizontal segment (1, 5) to (8, 5) has d = 7 and slope 0.

Worked example - 3D (0, 0, 0) to (1, 2, 2):

d = √(1 + 4 + 4) = √9 = 3. Midpoint (0.5, 1, 1). No slope line is printed in 3D mode because a space segment does not have a single rise/run in one plane unless you pick a projection.

Distance is a length ≥ 0. Slope can be negative (downhill left to right), zero (horizontal), or undefined (vertical). Midpoint is a point, not a length: (1.5, 2) is not “1.5 units of distance.” People mix these three because they all come from the same two ordered pairs.

Average each coordinate independently: M = ((x1+x2)/2, (y1+y2)/2) in the plane, and the same with z in space. For the default, M = (1.5, 2). The midpoint is the unique point on the segment equally far from both ends in Euclidean distance. It is also the centroid of a two-point set.

Section formula for a ratio other than 1:1 (dividing a segment in 2:1, for example) is not implemented. Neither is a weighted midpoint. If you need the perpendicular bisector, you already have M and, in 2D, the slope of the segment; the bisector’s slope is the negative reciprocal when the original slope exists and is nonzero - compute that on paper or with the slope tool.

• Forgetting to square before adding or forgetting the square root (reporting 9+16=25 as the distance instead of 5).
• Subtracting in the wrong order - signs of Δx and Δy flip, but squares make distance the same. Slope would flip if you swapped the points; this tool always uses point 2 minus point 1 as labeled.
• Treating slope as distance (4/3 is not 5).
• Using 2D mode for a 3D problem and dropping z, which understates the space diagonal.
• Lat/long as if they were metres.

Write the two ordered pairs. Subtract x-coordinates to get the run Δx = x2 − x1. Subtract y-coordinates to get the rise Δy = y2 − y1. Square each difference - squaring discards the sign, which is why distance does not care which point you call “first” for the length (though this UI always subtracts point 2 minus point 1 so slope keeps a direction). Add the squares. Take the principal (non-negative) square root. That root is d.

Check with a right triangle you already know. If Δx = 5 and Δy = 12, d = √(25+144) = 13. If Δx = 8 and Δy = 15, d = 17. If both differences are 1, d = √2 ≈ 1.4142135623730951, which is not 1.4 exactly and not 2. Sketching the segment on graph paper and counting diagonal boxes is a count of √2 steps, not of taxicab blocks.

Order of points: distance((A),(B)) = distance((B),(A)). Midpoint is also symmetric. Slope flips sign if you swap the labels of the two points, because rise and run both flip. This calculator does not swap them for you; x1,y1 is the first point as typed.

A rectangular room that is 3 m long, 4 m wide, and 12 m high has a space diagonal from one lower corner to the opposite upper corner of √(9+16+144) = √169 = 13 m. That is the 3D distance formula with points (0,0,0) and (3,4,12). You can think of it as a 5 m floor diagonal (3-4-5) then a right triangle of legs 5 and 12, hypotenuse 13. The same stacking of Pythagoras is what the single square root with three terms is doing in one shot.

In 3D there is no single “slope of the segment” without choosing a plane. Direction is a unit vector (Δx, Δy, Δz)/d, which this page does not print. Polar/azimuth angles are also omitted. If you needed the shadow of the segment on the xy-plane, that shadow’s length is the 2D distance with z ignored: √(Δx²+Δy²), which is shorter than the space diagonal unless Δz = 0.

Negative z is allowed. From (0, 0, 2) to (0, 0, −3) the distance is 5, a vertical drop. Midpoint is (0, 0, −0.5). Forgetting the sign of z and typing both as positive would give distance 1, which is a different physical pair of points.

Translating both points by the same vector does not change d: (0,0) to (3,4) is the same length as (10,10) to (13,14). Rotating the axes does not change d either; Euclidean distance is invariant under rigid motions. That is why the formula only sees differences. An origin far from the segment does not inflate the length unless you mistakenly use the coordinates themselves instead of the differences - √(x2²+y2²) is the distance from the origin to point 2, not from point 1 to point 2, unless point 1 is the origin.

Scaling the whole figure by k multiplies d by |k|. Reflecting across an axis leaves d alone. Those facts are why a CAD block moved around a drawing keeps its measured length, and why a mirrored floor plan still has the same room diagonals.

Taxicab (Manhattan) distance is |Δx| + |Δy|, the length of an axis-aligned path. For the default (0,0) to (3,4) that is 7, not 5. Chessboard (Chebyshev) distance is max(|Δx|, |Δy|) = 4. Euclidean distance - this page - is the straight chord 5. Navigation apps that route on a street grid are closer to Manhattan plus one-way constraints; a bird or a laser is Euclidean. Mixing the three in a lab report will not match a key that used √(Δx²+Δy²).

In 3D the taxicab analogue is |Δx|+|Δy|+|Δz|. For (0,0,0) to (1,2,2) that is 5, while Euclidean is 3. This calculator never switches norms. If an assignment asked for “city block distance,” do not use this result.

• Euclidean only. No Manhattan, Chebyshev, or Minkowski p-norms other than p = 2.
• No map geodesics and no unit conversion.
• Slope only in 2D, undefined on a vertical line or coincident points.
• IEEE-754 square roots. Enormous coordinates can overflow before the root.
• Two points only. No polyline length, no nearest-point-on-a-line.