Euclidean Distance Between Two 2D Points Calculator
The Cartesian distance formula applies the Pythagorean theorem to horizontal displacement Δx = |x₂ - x₁| and vertical displacement Δy = |y₂ - y₁| to find the Euclidean metric d = √(Δx² + Δy²).
Straight-line vector magnitude between points
Horizontal coordinate difference
Vertical coordinate difference
Step-by-Step Calculation
The Cartesian distance formula applies the Pythagorean theorem to horizontal displacement Δx = |x₂ - x₁| and vertical displacement Δy = |y₂ - y₁| to find the Euclidean metric d = √(Δx² + Δy²).
Real-World Reference Values
Shortest straight-line geodesic in flat space
Grid street distance along orthogonal axes
At Δx = 6, Δy = 8, slope m = 1.333 (53.13°)
20-Row Comparison Table
Euclidean Distance (d) · Horizontal Delta |x₂ - x₁| (Δx)
| # | Horizontal Delta |x₂ - x₁| (m) | Euclidean Distance (m) | Δ % |
|---|---|---|---|
| 01 | 6 mCurrent | 10 m | 0.00% |
| 02 | 724 m | 724.0442 m | +7140.44% |
| 03 | 1,448.00 m | 1,448.02 m | +14380.22% |
| 04 | 2,172.00 m | 2,172.01 m | +21620.15% |
| 05 | 2,896.00 m | 2,896.01 m | +28860.11% |
| 06 | 3,620.00 m | 3,620.01 m | +36100.09% |
| 07 | 4,344.00 m | 4,344.01 m | +43340.07% |
| 08 | 5,068.00 m | 5,068.01 m | +50580.06% |
| 09 | 5,792.00 m | 5,792.01 m | +57820.06% |
| 10 | 6,516.00 m | 6,516.00 m | +65060.05% |
| 11 | 7,240.00 m | 7,240.00 m | +72300.04% |
| 12 | 7,964.00 m | 7,964.00 m | +79540.04% |
| 13 | 8,688.00 m | 8,688.00 m | +86780.04% |
| 14 | 9,412.00 m | 9,412.00 m | +94020.03% |
| 15 | 10,136.00 m | 10,136.00 m | +101260.03% |
| 16 | 10,860.00 m | 10,860.00 m | +108500.03% |
| 17 | 11,584.00 m | 11,584.00 m | +115740.03% |
| 18 | 12,308.00 m | 12,308.00 m | +122980.03% |
| 19 | 13,032.00 m | 13,032.00 m | +130220.02% |
| 20 | 13,756.00 m | 13,756.00 m | +137460.02% |
Frequently Asked Questions
How do I find the horizontal span Δx given total distance d and vertical rise Δy?+
Click SOLVE FOR on Δx to evaluate Δx = √(d² - Δy²).
Does the sign of (x₂ - x₁) affect Euclidean distance?+
No. Squaring Δx and Δy makes both terms non-negative regardless of quadrant direction.
How does this extend to 3D coordinates?+
In 3D space, add the squared z-axis difference: d = √(Δx² + Δy² + Δz²).