Distance Formula Calculator

The distance between two points in the Euclidean plane is one of basic concepts in Geometry. Although, it is not a static or universal concept, as there many potential measures of "distance" in Math.

Indeed, different types of geometry can use different types of distances. And all of those geometries, including the Euclidean geometry, all define distances that are logical and consistent, and hold all the properties that are expected for a distance.

This calculator is based on the distance for the Euclidean geometry. Assume that we have two points \((x_1, y_1)\) and \((x_2, y_2)\), then the distance formula is computed as follows:

\[ D = \displaystyle \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]

Explanation

The expression above defines how to use the formula for the given two points. What is done is simple: the first component of point 1 and the first component of point 2 are subtracted, and the result is squared.

The same is done for the second point: the second component of point 1 and the second component of point 2 are subtracted, and the result is squared. These two squared values are added up, and you take square root to the result of it. The final number you obtain is the distance

Examples

Assume that we have two points \((1, 3)\) and \((4, 8)\), then the distance formula is computed as follows:

\[ D = \displaystyle \sqrt{(1 - 4)^2 + (3 - 8)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \]

The square root above \(\sqrt 34\) cannot be simplified any further, so we leave it like that. Sometimes, you will asked to deliver an approximate decimal answer, which in this case would be \(\sqrt 34 \approx 5.8310 \).

More Examples

How to handle the distance formula with fractions? It is all the same mechanic. Assume that we have two points \((\frac{1}{2}, \frac{1}{4})\) and \((\frac{3}{5}, \frac{3}{4})\), then the distance formula is computed as follows:

\[ D = \displaystyle \sqrt{ \left(\frac{1}{2} - \frac{3}{5}\right)^2 + \left(\frac{1}{4} - \frac{3}{4}\right)^2} = \sqrt{\left(-\frac{1}{10}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{1}{100} + \frac{1}{4}} = \sqrt{\frac{13}{50}} \approx 5.8310 \]

Does it have to be in two dimensions?

Not necessarily. Actually we can have two points in a n-dimensional space: \(u = (u_1, u_2, ..., u_n)\) and \(v = (v_1, v_2, ..., v_n)\). The distance is now calculated by squaring the differences of all the components, adding them up and taking square root:

\[ D = \displaystyle \sqrt{(u_1 - v_1)^2 + (u_2 - v_2)^2 + ... + + (u_n - v_n)^2} \]

Does this have to do anything with Pythagoras

You bet it does! As your intuition is correctly telling you, the square root of the sum of squares resembles a lot that of the Pythagorean theorem and also what you do when you solve triangles.

This is because we are defining the distance between two points in the Pythagorean way of geometry, as the size of the hypotenuse for a triangle in which the vertices are defined by the points given.

Or alternatively, you can get those two points and compute the midpoint .