Distance between point and the line Part 2

After I published my first solution I got feedback that the same can be achieved calculating other way. So I tried implement another way:

public double distanceBetweenLineAndPoints(Point a, Point b, Point c){
    double dx = a.x - b.x;
    double dy = a.y - b.y;
    if(dy == 0){
        //we have vertical line
        return distanceBetweenPoints(new Point(c.x, a.y), c);
    }
    if(dx == 0){
        //we have horizontal line
        return distanceBetweenPoints(new Point(a.x, c.y), c);
    }
    double slope = dy / dx;
    //lets check - may be c point is on the line
    if(c.y - a.y == slope * (c.x - a.x)){
        //yes, it is
        return 0;
    }
    Point aa = new Point();
    Point bb = new Point();
    aa.x = c.x;
    aa.y = slope * (c.x - a.x)+ a.y;
    bb.x = (c.y - a.y)/slope + a.x;
    bb.y = c.y;

    return (distanceBetweenPoints(bb, c) / distanceBetweenPoints(aa, bb)) * 
   distanceBetweenPoints(aa, c);
}

Is it faster?

I did two tests:
1. I took solution 1 and 2 gave them the same data (line is not horizontal/vertical and point is not on the line) - solution 1 was 10-15% faster.
2. I add the same checking (the line is vertical/horizontal or point is on the line) to solution 1 and repeated test - both solution were performing almost identically.