Math.hypot()

Baseline Widely available

This feature is well established and works across many devices and browser versions. It’s been available across browsers since ⁨July 2015⁩.

TheMath.hypot() static method returns the square root of the sum of squares of its arguments. That is,

𝙼𝚊𝚝𝚑.𝚑𝚢𝚙𝚘𝚝 (v_{1}, v_{2}, \dots, v_{n}) = \sqrt{\sum_{i = 1}^{n} v_{i}^{2}} = \sqrt{v_{1}^{2} + v_{2}^{2} + \dots + v_{n}^{2}} \mathtt{\operatorname{Math.hypot}(v_1, v_2, \dots, v_n)} = \sqrt{\sum_{i=1}^n v_i^2} = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2}

Try it

console.log(Math.hypot(3, 4));// Expected output: 5console.log(Math.hypot(5, 12));// Expected output: 13console.log(Math.hypot(3, 4, 5));// Expected output: 7.0710678118654755console.log(Math.hypot(-5));// Expected output: 5

The square root of the sum of squares of the given arguments. ReturnsInfinity if any of the arguments is ±Infinity. Otherwise, if at least one of the arguments is or is converted toNaN, returnsNaN. Returns0 if no arguments are given or all arguments are ±0.

Description

Calculating the hypotenuse of a right triangle, or the magnitude of a complex number, uses the formulaMath.sqrt(v1*v1 + v2*v2), where v1 and v2 are the lengths of the triangle's legs, or the complex number's real and complex components. The corresponding distance in 2 or more dimensions can be calculated by adding more squares under the square root:Math.sqrt(v1*v1 + v2*v2 + v3*v3 + v4*v4).

This function makes this calculation easier and faster; you callMath.hypot(v1, v2), orMath.hypot(v1, /* …, */, vN).

Math.hypot also avoids overflow/underflow problems if the magnitude of your numbers is very large. The largest number you can represent in JS isNumber.MAX_VALUE, which is around 10³⁰⁸. If your numbers are larger than about 10¹⁵⁴, taking the square of them will result in Infinity. For example,Math.sqrt(1e200*1e200 + 1e200*1e200) = Infinity. If you usehypot() instead, you get a better answer:Math.hypot(1e200, 1e200) = 1.4142...e+200. This is also true with very small numbers.Math.sqrt(1e-200*1e-200 + 1e-200*1e-200) = 0, butMath.hypot(1e-200, 1e-200) = 1.4142...e-200.

With one argument,Math.hypot() is equivalent toMath.abs().Math.hypot.length is 2, which weakly signals that it's designed to handle at least two parameters.

Becausehypot() is a static method ofMath, you always use it asMath.hypot(), rather than as a method of aMath object you created (Math is not a constructor).

Examples

Using Math.hypot()

Math.hypot(3, 4); // 5Math.hypot(3, 4, 5); // 7.0710678118654755Math.hypot(); // 0Math.hypot(NaN); // NaNMath.hypot(NaN, Infinity); // InfinityMath.hypot(3, 4, "foo"); // NaN, since +'foo' => NaNMath.hypot(3, 4, "5"); // 7.0710678118654755, +'5' => 5Math.hypot(-3); // 3, the same as Math.abs(-3)