Understanding battery hough transform

I don't know if I am getting things right or something else. This is what the transformation of the charts looks like:

  You just
  loop through each pixel and calculate rho and theta with this formula: r = x.cos (θ) + y.sin (θ)
  Most people put this formula in a loop of 0-> 180 or 0-360.
  What for?
  after that you put all values ​​into accumulator.

But for some dark and mysterious reason, all points on the same line will give the same values ​​for theta and r. How is this possible? I really do not understand.

I tried it, these were my results: I just put a straight black horizontal line on the image. The line was 10px long by 1px high. So I applied the formula, but instead of getting 1 place in my accumulator with a value of 10. I got 5 places with a value of 10, why?

this is my code if needed:

int outputImage[400][400];
int accumulator[5000][5000]; //random size
int i,j; 
int r,t;
int x1,y1,x2,y2;

/* PUT ALL VALUES IN ACCU */
         for(i=0;i<imageSize[0];i++)  
            {  
                 for( j=0;j<imageSize[1];j++)  
                 {  
                    if (inputImage[i][j] == 0x00) //just to test everything, formula only applied on the black line
                          { 

                           for( t=0;t<180;t++)  
                           { 
                              r = i  * cos(t) + j*sin(t);
                               accumulator[r][t] ++;                      
                           }   
                     }
                 }  
           }  


/*READ ACCU, and draw lines */
       for (i=0; i<5000;i++)
      {
        for (j=0;j<5000;j++)
        {
           if(accumulator[i][j] >= 10) // If accu value >= tresholdvalue, I believe we have a line
            {
              x1 = accumulator[i][j]  * i / (cos(accumulator[i][j] * j));
              y1 = 0;
              x2 = 0;
              y2 =  accumulator[i][j]  * i / (sin(accumulator[i][j] * j));
           // now i just need to draw all the lines between x1y1 and x2y2
            }
        }
       }

      

        
        
        
      

    

Thank!