}
+double Geometry::DistanceToLineFromPoint(Point tail, Point head, Point point)
+{
+ // Interpretation: given a line in the form x = a + tu, where u is the
+ // unit vector of the line, a is the tail and t is a parameter which
+ // describes the line, the distance of a point p to the line is given by:
+ // || (a - p) - ((a - p) . u) u ||
+ // We go an extra step: we set the sign to reflect which side of the line
+ // it's on (+ == to the left if head points away from you, - == to the
+ // right)
+ Vector line(tail, head);
+ Vector u = line.Unit();
+ Vector a_p = tail - point;
+ Vector dist = a_p - (u * (a_p).Dot(u));
+
+ double angle = Vector::Angle(tail, point) - line.Angle();
+
+ if (angle < 0)
+ angle += TAU;
+
+ return dist.Magnitude() * (angle < HALF_TAU ? +1.0 : -1.0);
+}
+
+
Point Geometry::MirrorPointAroundLine(Point point, Point tail, Point head)
{
// Get the vector of the intersection of the line and the normal on the
// Radius is the hypotenuse, so we have to use c² = a² + b² => a² = c² - b²
double perpendicularLength = sqrt((c->radius[0] * c->radius[0]) - (distance * distance));
- // Now, find the points using the length, then check to see if they are on
- // the line segment
+ // Now, find the intersection points using the length...
Vector lineUnit = Vector(l->p[0], l->p[1]).Unit();
Point i1 = p + (lineUnit * perpendicularLength);
Point i2 = p - (lineUnit * perpendicularLength);
- // Now we have our intersection points, next we need to see if they are on
- // the line segment...
+ // Next we need to see if they are on the line segment...
double u = ParameterOfLineAndPoint(l->p[0], l->p[1], i1);
double v = ParameterOfLineAndPoint(l->p[0], l->p[1], i2);
return Point(0, 0);
}
+
+Point Geometry::Midpoint(Line * line)
+{
+ return Point((line->p[0].x + line->p[1].x) / 2.0,
+ (line->p[0].y + line->p[1].y) / 2.0);
+}
+
+
+/*
+How to find the tangent of a point off a circle:
+
+ • Calculate the midpoint on the point and the center of the circle
+ • Get the length of the line segment from the and the center divided by two
+ • Use that length to construct a circle with the point at the center and the
+ radius equal to that length
+ • The intersection of the two circles are the tangent points
+
+*/
+