Free Distance Between Two Points Calculator
Compute the straight-line distance and the midpoint between any two points in the coordinate plane, with the full distance-formula arithmetic shown for each result.
Enter the coordinates of two points to find the distance and midpoint.
d = √((x₂ − x₁)² + (y₂ − y₁)²) = √(3² + 4²)
Quick answer
The distance between two points is d = √((x2−x1)² + (y2−y1)²), the length of the straight line joining them. For the points (0,0) and (3,4): d = √(3² + 4²) = √(9 + 16) = √25 = 5, and the midpoint is ((0+3)/2, (0+4)/2) = (1.5, 2).
Formula & method
Distance (Euclidean)
d = √((x₂ − x₁)² + (y₂ − y₁)²)
- d — straight-line distance between the two points
- x₁, y₁ — coordinates of the first point
- x₂, y₂ — coordinates of the second point
The 2-D Euclidean distance — a direct application of the Pythagorean theorem, where the horizontal gap (x₂−x₁) and vertical gap (y₂−y₁) are the legs of a right triangle and d is the hypotenuse.
Midpoint
M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)
The midpoint is the point exactly halfway along the segment; each coordinate is the average of the corresponding coordinates of the endpoints.
Examples
- Input
- (x₁,y₁) = (0,0), (x₂,y₂) = (3,4)
- Result
- d = 5, midpoint = (1.5, 2)
- Why
- Δx = 3−0 = 3 and Δy = 4−0 = 4, so d = √(3² + 4²) = √(9 + 16) = √25 = 5. Midpoint = ((0+3)/2, (0+4)/2) = (1.5, 2).
- Input
- (x₁,y₁) = (1,2), (x₂,y₂) = (4,6)
- Result
- d = 5, midpoint = (2.5, 4)
- Why
- Δx = 4−1 = 3 and Δy = 6−2 = 4, so d = √(3² + 4²) = √(9 + 16) = √25 = 5. Midpoint = ((1+4)/2, (2+6)/2) = (2.5, 4).
- Input
- (x₁,y₁) = (−2,−3), (x₂,y₂) = (4,5)
- Result
- d = 10, midpoint = (1, 1)
- Why
- Δx = 4−(−2) = 6 and Δy = 5−(−3) = 8, so d = √(6² + 8²) = √(36 + 64) = √100 = 10. Midpoint = ((−2+4)/2, (−3+5)/2) = (1, 1).
- Input
- (x₁,y₁) = (2,3), (x₂,y₂) = (2,9)
- Result
- d = 6, midpoint = (2, 6)
- Why
- Δx = 2−2 = 0 and Δy = 9−3 = 6, so d = √(0² + 6²) = √36 = 6 — the distance is just the difference in y. Midpoint = ((2+2)/2, (3+9)/2) = (2, 6).
When to use this tool
- Measuring the straight-line (as-the-crow-flies) distance between two map or screen coordinates.
- Geometry and algebra homework: verifying segment lengths, midpoints, or whether three points are collinear or form a right triangle.
- Game development, graphics, and physics, where you need the gap between two (x,y) positions for collision, range, or movement checks.
- Surveying, CAD, and engineering layout when working from Cartesian coordinates on a single flat plane.
Common mistakes
- Forgetting to square the differences before adding — you must compute (Δx)² + (Δy)², not (Δx + Δy). For (3,4) that gives √25 = 5, not √7.
- Mishandling negative coordinates: subtracting a negative is adding. With x₁ = −2 and x₂ = 4, Δx = 4 − (−2) = 6, not 2.
- Taking the square root of each term separately. √(a² + b²) ≠ a + b; the square root applies to the whole sum, so the radical must come last.
- Confusing the midpoint formula (average the coordinates) with the distance formula, or dividing the distance by 2 to get the midpoint — the midpoint is a point, not a length.
Frequently asked questions
What is the distance formula?
It is d = √((x₂ − x₁)² + (y₂ − y₁)²). It gives the straight-line distance between two points in the plane and comes directly from the Pythagorean theorem, treating the horizontal and vertical gaps as the two legs of a right triangle.
Does the order of the two points matter?
No. Because the differences are squared, (x₂ − x₁)² equals (x₁ − x₂)², so swapping which point is first gives the same distance. The midpoint is also unaffected, since addition is commutative.
How do I find the midpoint between two points?
Average the x-coordinates and average the y-coordinates: M = ((x₁ + x₂)/2, (y₁ + y₂)/2). For (0,0) and (3,4) the midpoint is (1.5, 2). The midpoint always lies exactly halfway along the segment.
Can I use negative coordinates?
Yes. Just substitute the signed values and be careful with subtraction: 4 − (−2) = 6. The formula works in every quadrant, and the resulting distance is always zero or positive.
Why is the distance formula the same as the Pythagorean theorem?
Draw a right triangle whose horizontal leg is |x₂ − x₁| and whose vertical leg is |y₂ − y₁|. The segment between the points is the hypotenuse, so by a² + b² = c² the distance is √((x₂−x₁)² + (y₂−y₁)²).
Does this work for 3-D points or latitude/longitude?
This calculator is for 2-D Cartesian points. For 3-D you add a (z₂ − z₁)² term inside the square root. Latitude/longitude needs the great-circle (haversine) formula because the Earth is curved, not flat.
Sources & references
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