The Math.hypot() function returns the square root of the sum of squares of its arguments, that is:

Math.hypot(v1,v2,,vn)=i=1nvi2=v12+v22++vn2\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}


Math.hypot([value1[, value2[, ...]]])


value1, value2, ...

Return value

The square root of the sum of squares of the given arguments. If at least one of the arguments cannot be converted to a number, NaN is returned.


Calculating the hypotenuse of a right triangle, or the magnitude of a complex number, uses the formula Math.sqrt(v1*v1 + v2*v2) where v1 and v2 are either the sides of the triangle, or the real and complex values.  For calculating distance in 2 or more dimensions, simply add in more squares inside the square root sign, like Math.sqrt(v1*v1 + v2*v2 + v3*v3 + v4*v4).  

This function makes it a little easier and faster, you just call Math.hypot(v1, v2) , or Math.hypot(v1, v2, v3, v4, ...) .  

It also avoids a problem if the magnitude of your numbers is huge.  The largest number you can represent in JS's double floats is Number.MAX_VALUE = 1.797...e+308 .  If your numbers are larger  than about 1e154, taking the square of them will result in Infinity, demolishing your results.  For example, Math.sqrt(1e200*1e200 + 1e200*1e200) = Infinity .  If you use hypot() instead, you get a good 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, but Math.hypot(1e-200, 1e-200) =1.4142...e-200, a good answer.

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

If no arguments are given, the result is +0.

If at least one of the arguments cannot be converted to a number, the result is NaN.

With one argument, Math.hypot() returns the same as Math.abs().


This can be emulated using the following function:

if (!Math.hypot) Math.hypot = function() {
  var y = 0, i = arguments.length;
  while (i--) y += arguments[i] * arguments[i];
  return Math.sqrt(y);

A polyfill that avoids underflows and overflows:

if (!Math.hypot) Math.hypot = function (x, y) {
  var max = 0;
  var s = 0;
  for (var i = 0; i < arguments.length; i += 1) {
    var arg = Math.abs(Number(arguments[i]));
    if (arg > max) {
      s *= (max / arg) * (max / arg);
      max = arg;
    s += arg === 0 && max === 0 ? 0 : (arg / max) * (arg / max);
  return max === 1 / 0 ? 1 / 0 : max * Math.sqrt(s);


Using Math.hypot()

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


ECMAScript (ECMA-262)
The definition of 'Math.hypot' in that specification.

Browser compatibility

Update compatibility data on GitHub
ChromeEdgeFirefoxInternet ExplorerOperaSafariAndroid webviewChrome for AndroidFirefox for AndroidOpera for AndroidSafari on iOSSamsung InternetNode.js
hypotChrome Full support 38Edge Full support 12Firefox Full support 27IE No support NoOpera Full support 25Safari Full support 8WebView Android Full support 38Chrome Android Full support 38Firefox Android Full support 27Opera Android Full support 25Safari iOS Full support 8Samsung Internet Android Full support 3.0nodejs Full support 0.12


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See also