Finding which side one point is from another point

Question

I have a simple question I think, if point a is to the right of another point I a dirx to be 1, otherwise -1. Likewise for a diry var, if it's above the other point I want it to be -1 otherwise 1. This is what I've worked out so far:

dirx = (abs(disx+1)/(disx+1))*-1;
diry = (abs(disy+1)/(disy+1))*-1;

The disx and disy are the distance from each other on x and y. I have a +1 to keep from having a division by 0 error, lastly I'm *-1 it all so that it'll match the coordinate system of the game screen where something higher has a lesser value than something lower.

My problem is that I'm still getting a division by 0 error with this code, I'm not sure why as I even added +1 to make sure it wasn't possible???

Answer 1

if disX = -1, then you'll always get a divide by zero error using that code.

If you're programming in a c-based language, just use the ? operator, a la:

if(disX == 0)
{  
  dirX = 0;  
}
else
{  
  dirx = (disX > 0) ? 1 : -1;  
}

Answer 2

If I understood your question correctly you just wanted to normalize a vector.

let me break your question into 3 parts :

1. why do you get division by zero?
   it's simply because either disx or disy values could be exactly (-1). in that situation you computer will start computing (abs(-1+1)/(-1+1))*-1 which is basically (0/0)*-1!

2. how to find sign of a number (or how to normalize 1D vector)?
   let's say you have a number 'X' and you want to check if this number is negative or positive. usually there is a function in math libraries named sign or sgn. They implement that function like this:

   int sign(int x)
   {
       if (x > 0)
           return 1;
           else
           if (x < 0)
               return -1;
           else
               return 0;
   }
   

   note that there is one extra condition where X is exactly zero, in that case sign function should return 0 since x doesn't have any sign!

3. how to normalize a 2D Vector (or a vector in higher dimensions)?
   the basic idea is to create a vector which has a length of exactly 1 so the first step is to calculate length of current vector.

   length = sqrt(vec.x^2 + vec.y^2 + vec.z^2 + ...)
   

   the next step is to shorten (or stretch) your vector so that it's total length equals to exactly 1. and here is how it's done:

   vec.x = vec.x / length;
   vec.y = vec.y / length;
   vec.z = vec.z / length;
   .
   .
   .
   

again note that length is always some positive value but it can be equal to zero (where all elements in your vector are exactly zero). you have to check if a vector's length is zero you can't compute a normalized vector!

Answer 3

What you need is the Cross Product of the two points and check if it is bigger or smaller than zero (zero means it is on the same line):

Point a Point b

cp = a.x*b.y - a.y*b.x;

For example:

a= (2,5); b= (3,4) cp = 2*4 - 5*3 = 8 - 15 = -7

as cp < 0 the point a is on the left of point b

Your function can thus just do this to return -1 or 1:

cp = a.x*b.y - a.y*b.x;
if(cp<0)
   return -1;
else
   return 1;