DistCalc

Distance Calculator

Distance formula: √((x₂−x₁)² + (y₂−y₁)²).

x₁

y₁

x₂

y₂

Distance

d = √((x₂-x₁)² + (y₂-y₁)²)

What you entered

x₁: 0y₁: 0x₂: 3y₂: 4

1
Δx (horizontal distance)
3 − 0= 3
2
Δy (vertical distance)
4 − 0= 4
3
Distance
√(3² + 4²)= 5

Result

Distance: 5

The distance between (0, 0) and (3, 4) is 5 units.

Distance between two points

The distance formula is Pythagoras applied to the horizontal and vertical gaps between the points.

Point 1	Point 2	Distance
(0, 0)	(3, 4)	5.0000
(2, 3)	(8, 3)	6.0000
(1, 2)	(5, 10)	8.9443
(0, 0)	(6, 8)	10.0000
(-2, -3)	(4, 5)	10.0000

Rows one and four are the 3-4-5 triangle and its double, giving exact whole numbers - most coordinate pairs do not, as row three shows. The last row spans negative coordinates and still gives exactly 10, because the differences are 6 and 8 regardless of sign: squaring removes the sign, so the order of the two points never matters either. Row two is horizontal, where the distance is simply the difference in x.

Pythagoras in disguise

The distance formula is just the Pythagorean theorem applied to coordinates — the horizontal and vertical gaps between two points form the legs of a right triangle, and the straight-line distance is its hypotenuse.

Where it's used

Mapping applications, video game physics, and any calculation of 'as the crow flies' distance between two locations on a grid or coordinate plane use exactly this formula.

Frequently asked questions

How do I find the distance between two points?

d = √((x₂−x₁)² + (y₂−y₁)²). For (1, 2) and (4, 6): d = √((3)² + (4)²) = √(9 + 16) = √25 = 5. This is the straight-line (Euclidean) distance, not the walking distance along streets.

Does the distance formula work in 3D?

Yes — add a z-component: d = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²). For points (1,2,3) and (4,6,3): d = √(9 + 16 + 0) = 5. Same Pythagorean idea, extended to three dimensions.

How is the distance formula related to the Pythagorean theorem?

They're the same thing. The horizontal distance (x₂−x₁) and vertical distance (y₂−y₁) form the two legs of a right triangle. The straight-line distance between the points is the hypotenuse: c = √(a² + b²).