Merge pull request #24216 from dkurt:inter_lines_less_compute

Minor optimization of two lines intersection #24216

### Pull Request Readiness Checklist

Not significant, but we can reduce number of multiplications while compute two lines intersection. Both methods are used heavily in their modules.

See details at https://github.com/opencv/opencv/wiki/How_to_contribute#making-a-good-pull-request

- [x] I agree to contribute to the project under Apache 2 License.
- [x] To the best of my knowledge, the proposed patch is not based on a code under GPL or another license that is incompatible with OpenCV
- [x] The PR is proposed to the proper branch
- [ ] There is a reference to the original bug report and related work
- [x] There is accuracy test, performance test and test data in opencv_extra repository, if applicable
      Patch to opencv_extra has the same branch name.
- [x] The feature is well documented and sample code can be built with the project CMake

modules/imgproc/src/geometry.cpp

@@ -328,17 +328,16 @@ static LineSegmentIntersection parallelInt( Point2f a, Point2f b, Point2f c, Poi
 static LineSegmentIntersection intersectLineSegments( Point2f a, Point2f b, Point2f c,
                                                       Point2f d, Point2f& p, Point2f& q )
 {
-    double denom = a.x * (double)(d.y - c.y) + b.x * (double)(c.y - d.y) +
-                   d.x * (double)(b.y - a.y) + c.x * (double)(a.y - b.y);
+    double denom = (a.x - b.x) * (double)(d.y - c.y) - (a.y - b.y) * (double)(d.x - c.x);
 
     // If denom is zero, then segments are parallel: handle separately.
     if( denom == 0. )
         return parallelInt(a, b, c, d, p, q);
 
-    double num = a.x * (double)(d.y - c.y) + c.x * (double)(a.y - d.y) + d.x * (double)(c.y - a.y);
+    double num = (d.y - a.y) * (double)(a.x - c.x) + (a.x - d.x) * (double)(a.y - c.y);
     double s = num / denom;
 
-    num = a.x * (double)(b.y - c.y) + b.x * (double)(c.y - a.y) + c.x * (double)(a.y - b.y);
+    num = (b.y - a.y) * (double)(a.x - c.x) + (c.y - a.y) * (double)(b.x - a.x);
     double t = num / denom;
 
     p.x = (float)(a.x + s*(b.x - a.x));

modules/objdetect/src/qrcode.cpp

@@ -68,19 +68,14 @@ static void updatePointsResult(OutputArray points_, const vector<Point2f>& point
 
 static Point2f intersectionLines(Point2f a1, Point2f a2, Point2f b1, Point2f b2)
 {
+    // Try to solve a two lines intersection (a1, a2) and (b1, b2) as a system of equations:
+    // a2 + u * (a1 - a2) = b2 + v * (b1 - b2)
     const float divisor = (a1.x - a2.x) * (b1.y - b2.y) - (a1.y - a2.y) * (b1.x - b2.x);
     const float eps = 0.001f;
     if (abs(divisor) < eps)
         return a2;
-    Point2f result_square_angle(
-                              ((a1.x * a2.y  -  a1.y * a2.x) * (b1.x - b2.x) -
-                               (b1.x * b2.y  -  b1.y * b2.x) * (a1.x - a2.x)) /
-                               divisor,
-                              ((a1.x * a2.y  -  a1.y * a2.x) * (b1.y - b2.y) -
-                               (b1.x * b2.y  -  b1.y * b2.x) * (a1.y - a2.y)) /
-                               divisor
-                              );
-    return result_square_angle;
+    const float u = ((b2.x - a2.x) * (b1.y - b2.y) + (b1.x - b2.x) * (a2.y - b2.y)) / divisor;
+    return a2 + u * (a1 - a2);
 }
 
 //      / | b