Count Pixel

edit

You should use LineIterator class: http://docs.opencv.org/modules/core/doc/drawing_functions.html?highlight=lineiterator#LineIterator

E.g. if you want to get pixel values failing on that line:

LineIterator it(img, pt1, pt2, 8);
LineIterator it2 = it;
vector<Vec3b> buf(it.count);

for(int i = 0; i < it.count; i++, ++it)
    buf[i] = *(const Vec3b)*it;

it.count in this sample is pixels count for the line between pt1 and pt2 if we use 8 neighbors.

I do not think he wants to calculate the euclidean distance, but he wants to define the pixel values falling on that line.

You should first consider how a line defined by 2 points can be represented mathematically. Given point 1 and point 2, the line equation equals:

Point1 = Point(x1,y1); Point2 = Point(x2,y2);   
y = y1 + [(y2 - y1) / (x2 - x1)]·(x - x1);

This gives you the ability to know exactly which pixels correspond to that line. Loop over your image, enter the x coördinate in the equation and look if the output is still the index your are currently in y position. If this check holds, than add the current value of the pixel to a storage container, for example an int that increases in value each time.

for a straight line, the euklidean distance is:

sqrt( (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y) );