Note that, sqrt( (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) ) is the same as saying sqrt( (a - b).x * (a - b).x + (a - b).y * (a - b).y ), so you can call cv::norm(a - b).

However, according to this (albeit old) link, performance is poor for single point pairs.

I just ran the test code on my machine. It generates 15,000 points and calculates the distance from each to the remaining points.

a : [0, 0] - b : [2.14748e+09, 2.14748e+09]

euclideanDist : 3.037e+09
distanceBtwPoints : 3.037e+09
cv::norm : 3.037e+09

max_distance euclideanDist : 3.02456e+09 time passed :0.165179
max_distance distanceBtwPoints : 3.02456e+09 time passed :0.259471
max_distance cv::norm : 3.02456e+09 time passed :0.26728

Surprising. The fastest code is

float euclideanDist(cv::Point2f& a, cv::Point2f& b)
{
    cv::Point2f diff = a - b;
    return cv::sqrt(diff.x*diff.x + diff.y*diff.y);
}

Using cv::norm(a - b) and the code below are nearly equal:

static double distanceBtwPoints(const cv::Point2f &a, const cv::Point2f &b)
{
    double xDiff = a.x - b.x;
    double yDiff = a.y - b.y;

    return std::sqrt((xDiff * xDiff) + (yDiff * yDiff));
}

But that's evidently because of the casts to double. If left as float, distanceBtwPoints is as fast as eucledianDist.