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Distance Formula Calculator

Calculate the distance between two points using d = \u221A((x\u2082\u2212x\u2081)\u00B2 + (y\u2082\u2212y\u2081)\u00B2). Based on the Pythagorean theorem applied to coordinate geometry.

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Try an Example

Pick a scenario to see how the calculator works, then adjust the values

Ladder Against Wall

A 10-foot ladder leans against a wall with the base 6 feet away. How high does it reach?

Key values: Leg a = 6 ft · Hypotenuse c = 10 ft · Solve for leg b

TV Screen Diagonal

Find the diagonal of a TV with a 16:9 aspect ratio (width 40", height 22.5").

Key values: Width = 40 in · Height = 22.5 in · Find diagonal

Classic 3-4-5 Triangle

Verify that sides 3, 4, and 5 form a right triangle (the most famous Pythagorean triple).

Key values: Sides: 3, 4, 5 · Right triangle · Primitive triple

Documentation

The Distance Formula

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in the Cartesian plane is derived directly from the Pythagorean theorem:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Why It Works

Draw a right triangle with the two points as endpoints of the hypotenuse. The horizontal leg has length $|x_2 - x_1|$ and the vertical leg has length $|y_2 - y_1|$ . By the Pythagorean theorem, the hypotenuse (the distance) is $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ .

Key insight: The distance formula is not a separate concept to memorize—it is the Pythagorean theorem applied to coordinates. Every distance calculation in coordinate geometry is a right triangle in disguise.

Extension to Three Dimensions

The same principle extends naturally to 3D by applying the Pythagorean theorem twice:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

First compute the distance in the xy-plane, then treat that result and the z-difference as legs of a second right triangle. This generalizes to any number of dimensions:

d = \sqrt{\sum_{i=1}^{n} (p_i - q_i)^2}

This n-dimensional version is called the Euclidean distance and is fundamental to machine learning, statistics, and physics.

Worked Examples

Example 1: Distance Between Two Cities on a Map

On a coordinate grid where 1 unit = 1 mile, City A is at (3, 7) and City B is at (11, 1).

d = \sqrt{(11-3)^2 + (1-7)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ miles}

Example 2: Checking If a Point Lies on a Circle

Does the point (5, 12) lie on the circle centered at the origin with radius 13?

d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \quad \checkmark

Yes—the distance from the origin equals the radius, so the point is on the circle. (This is also the Pythagorean triple (5, 12, 13)!)

Example 3: Midpoint Verification

The midpoint of two points should be equidistant from both. For (2, 4) and (8, 10), the midpoint is (5, 7). Verify:

d_1 = \sqrt{(5-2)^2 + (7-4)^2} = \sqrt{9+9} = 3\sqrt{2}

d_2 = \sqrt{(8-5)^2 + (10-7)^2} = \sqrt{9+9} = 3\sqrt{2} \quad \checkmark

Applications

Navigation: GPS distance calculations between waypoints
Computer graphics: Collision detection, path length, object proximity
Machine learning: k-Nearest Neighbors (kNN), clustering algorithms (k-means), and dimensionality reduction all use Euclidean distance
Physics: Calculating displacement, electric field strength (inverse-square law), and gravitational force

Frequently Asked Questions

What is the distance formula?

The distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ calculates the straight-line distance between two points in a coordinate plane. It is derived directly from the Pythagorean theorem by treating the horizontal and vertical displacements as the legs of a right triangle.

How is the distance formula related to the Pythagorean theorem?

The distance formula is the Pythagorean theorem applied to coordinates. The horizontal displacement $|x_2 - x_1|$ and vertical displacement $|y_2 - y_1|$ form the two legs of a right triangle, and the distance between the points is the hypotenuse.

Does the distance formula work in three dimensions?

Yes. In 3D, add the z-component: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ . This generalizes to any number of dimensions as $d = \sqrt{\sum (\text{squared differences})}$ , known as the Euclidean distance.

Can I use the distance formula to check if a point is on a circle?

Yes. A point $(x, y)$ lies on a circle centered at $(h, k)$ with radius $r$ if the distance from the point to the center equals $r$ . Calculate $d = \sqrt{(x - h)^2 + (y - k)^2}$ and check if $d = r$ .

What are common applications of the distance formula?

Common applications include GPS navigation between waypoints, collision detection in computer graphics, k-nearest neighbor algorithms in machine learning, calculating displacement in physics, and verifying geometric properties like midpoint equidistance.

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