I want to calculate the distance between two points in Java

Question

OK, so I\'ve written most of a program that will allow me to determine if two circles overlap.

I have no problems whatsoever with my program aside from one issue: t

Answer 1

Math.sqrt returns a double so you'll have to cast it to int as well

distance = (int)Math.sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));

Answer 2

You could also you Point2D Java API class:

public static double distance(double x1, double y1, double x2, double y2)

Example:

double distance = Point2D.distance(3.0, 4.0, 5.0, 6.0);
System.out.println("The distance between the points is " + distance);

Answer 3

This may be OLD, but here is the best answer:

    float dist = (float) Math.sqrt(
            Math.pow(x1 - x2, 2) +
            Math.pow(y1 - y2, 2) );

Answer 4

Based on the @trashgod's comment, this is the simpliest way to calculate distance:

double distance = Math.hypot(x1-x2, y1-y2);

From documentation of Math.hypot:

Returns: sqrt(x²+ y²) without intermediate overflow or underflow.

Answer 5

Unlike maths-on-paper notation, most programming languages (Java included) need a * sign to do multiplication. Your distance calculation should therefore read:

distance = Math.sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));

Or alternatively:

distance = Math.sqrt(Math.pow((x1-x2), 2) + Math.pow((y1-y2), 2));

Answer 6

Based on the @trashgod's comment, this is the simpliest way to calculate >distance:

double distance = Math.hypot(x1-x2, y1-y2); From documentation of Math.hypot:

Returns: sqrt(x²+ y²) without intermediate overflow or underflow.

Bob

Below Bob's approved comment he said he couldn't explain what the

Math.hypot(x1-x2, y1-y2);

did. To explain a triangle has three sides. With two points you can find the length of those points based on the x,y of each. Xa=0, Ya=0 If thinking in Cartesian coordinates that is (0,0) and then Xb=5, Yb=9 Again, cartesian coordinates is (5,9). So if you were to plot those on a grid, the distance from from x to another x assuming they are on the same y axis is +5. and the distance along the Y axis from one to another assuming they are on the same x-axis is +9. (think number line) Thus one side of the triangle's length is 5, another side is 9. A hypotenuse is

 (x^2) + (y^2) = Hypotenuse^2

which is the length of the remaining side of a triangle. Thus being quite the same as a standard distance formula where

 Sqrt of (x1-x2)^2 + (y1-y2)^2 = distance 

because if you do away with the sqrt on the lefthand side of the operation and instead make distance^2 then you still have to get the sqrt from the distance. So the distance formula is the Pythagorean theorem but in a way that teachers can call it something different to confuse people.