判断点P在三角形内还是外

已经三角形ABC三个顶点的坐标，A(x1,y1),   B(x2,y2),  C(x3,y3) ,   和另外一点P(x,y) 的坐标，怎么判断在三角形内还是外
 
虽然做出来了，但是不高效，现在我也不知道高效的方法是什么？哪位大侠告诉小弟。
方法一：求面积之和
public class Mytest02 {
       public static void main(String[] args){
              double x1=1.0, y1=1.0 ;           //A
              double x2=3.0, y2=1.0 ;           //B
              double x3=2.0, y3=3.0 ;   //C
 
      
              double x=2.0, y=2.0 ;  //P
 
              double pa = Math.sqrt((x-x1)*(x-x1)+(y-y1)*(y-y1));
              double pb = Math.sqrt((x-x2)*(x-x2)+(y-y2)*(y-y2));
              double pc = Math.sqrt((x-x3)*(x-x3)+(y-y3)*(y-y3));
      
              double ab = Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
              double ac = Math.sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));
              double bc = Math.sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));
      
              double abc = (ab+bc+ac)/2 ;
      
              double pab = (pa+pb+ab)/2;
              double pbc = (pb+pc+bc)/2;
              double pac = (pa+pc+ac)/2;
System.out.println(abc);
System.out.println(pab);
System.out.println(pbc);
System.out.println(pac);
              double sabc = Math.sqrt(abc*(abc-ab)*(abc-bc)*(abc-ac));
      
              double spab = Math.sqrt(pab*(pab-pa)*(pab-pb)*(pab-ab));
              double spbc = Math.sqrt(pbc*(pbc-pb)*(pbc-pc)*(pbc-bc));
              double spac = Math.sqrt(pac*(pac-pa)*(pac-pc)*(pac-ac));
System.out.println(sabc);
System.out.println(spab);
System.out.println(spbc);
System.out.println(spac);
              //判断
              if(Math.abs(sabc-(spab+spbc+spac))<0.0000001){
                     System.out.println("P点在这个三角形内部或者边上");
              }else {
                     System.out.println("P点不在这个三角形外部");
              }
       }
 
   
}
 
 
方法二：求角度之和
public class Mytest03 {
       public static void main(String[] args){
              double x1=1.0, y1=1.0 ;           //A
              double x2=3.0, y2=1.0 ;           //B
              double x3=2.0, y3=3.0 ;   //C
 
      
              double x=2.0, y=2.0 ;  //P
 
              double pa = Math.sqrt((x-x1)*(x-x1)+(y-y1)*(y-y1));
              double pb = Math.sqrt((x-x2)*(x-x2)+(y-y2)*(y-y2));
              double pc = Math.sqrt((x-x3)*(x-x3)+(y-y3)*(y-y3));
      
              double ab = Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
              double ac = Math.sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));
              double bc = Math.sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));
      
double apb = Math.acos((pa*pa+pb*pb-ab*ab)/(2*pa*pb)) ;
double apc = Math.acos((pa*pa+pc*pc-ac*ac)/(2*pa*pc)) ;
double bpc = Math.acos((pb*pb+pc*pc-bc*bc)/(2*pb*pc)) ;
System.out.println(apb) ;
System.out.println(apc) ;
System.out.println(bpc) ;
System.out.println(Math.PI*2) ;
              //判断
              if(Math.abs(apb+apc+bpc-Math.PI*2)<0.0000001){
                     System.out.println("P点在这个三角形内部或者边上");
              }else {
                     System.out.println("P点在这个三角形外部");
              }
       }
 
}