汇总篇：

点到直线的距离 

cos(x)=BA*BC/(|BA|*|BC|）

求AD有很多种方法，可以用勾股定理

这里用的三角函数

x=acos(cos(x))

|AD|=|BA|*sin(x)

如果x是钝角，|AD|=|BA|*sin(pi-x)=|BA|*sin(x)

如果是直角，sin(x) = 1，|AD|=|BA|

点A到直线BC的投影点

方法一

设D(dx,dy)

AD=(dx-ax,dy-ay)

BC=(C.x-B.x,C.y-B.y)

|BA|=sqrt((bx-ax)^2+(by-ay)^2)

令AD*BC=0,(dx-ax)*(cx-bx)+(dy-ay)*(cy-by)=0

|AD|=sqrt((dx-ax)^2+(dy-ay)^2)= |BA|*sin(x)

解上述方程组可解得dx,dy

方法二

BA在BC上的投影等于BA乘以BC的方向向量

D = B +BD

#include<iostream>

class point{
 	public:
	double x;
	double y;
	point(double x_=0,double y_=0):x(x_),y(y_){} 
	friend const point operator+(const point& p1,const point& p2){
		return point(p1.x+p2.x,p1.y+p2.y);
	};
	friend const point operator-(const point& p1,const point& p2){
		return point(p1.x-p2.x,p1.y-p2.y);
	};
	friend const point operator*(const point& p,const double& m){
		return point(p.x*m,p.y*m);
	};
	friend const point operator*(const double& m,const point& p){
		return point(p.x*m,p.y*m);
	};
	friend const point operator/(const point& p,const double& m){
		return point(p.x/m,p.y/m);
	};
	friend ostream& operator <<(ostream& out,point& a){
		printf("(%lf,%lf)",a.x,a.y);
		return out;
	};
};
typedef point vect2;//重命名，向量也是用坐标表示 
 
class line{
	public:
	point start;
	point end; 
	line(point s=point(0,0),point e=point(0,0)):start(s),end(e){}
};

double dot(point O,point A,point B){//点乘 
	double oa_x=A.x-O.x;
	double oa_y=A.y-O.y;
	double ob_x=B.x-O.x;
	double ob_y=B.y-O.y;
	return oa_x*ob_x+oa_y*ob_y;
}
double dis(const point &p1,const point &p2){//求两点之间距离
	double ans=(p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);
	return sqrt(ans);
}
double dis2(const point &p1,const point &p2){//求两点之间距离的平方
	return(p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);	 
}
//点到直线的距离 
double disOfPointToLine(point O,line l){
	double cos0=dot(l.start,O,l.end)/(dis(O,l.start)*dis(l.start,l.end));
	return dis(O,l.start)*sin(acos(cos0));		
} 
//点在直线上的投影  
point shadowPointOfPointToLine(point A,line l){//投影点出了问题 
	point B = l.start;
	point C = l.end;
	point D;
	vect2 BC = C - B;
	vect2 BA = B - A;
	vect2 BD = BC * BA /dis2(B,C) * BC; 
	D = B + BD;
	return D;	
}