/*
思路分析
思路一 ： 
    1. 求两点之间的距离 l = sqrt(pow((x2 - x1) , 2) + pow((y2 - y1) , 2))
    2. 求点到线的距离 斜率k= (y2-y1)/(x2-x1) 截距b= y - kx (代入一个坐标进去)
        --- 距离d= | k * x3 - y3 + b | / sqrt(pow(k , 2) + 1)
    3. 底X高/2  s= l * d / 2;
思路二： 
    1. 求出三条边的距离
    3. 利用海伦公式直接求出 s = sqrt(p*(p-a)*(p-b)*(p-c)) p = (a + b + c) / 2

//思路一
#include <bits/stdc++.h>
using namespace std;

int main()
{
    double x1,y1,x2,y2,x3,y3;
    cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
    double l,d,s,k,b;
    l = sqrt(pow((x2 - x1) , 2) + pow((y2 - y1) , 2));
    k = (y2-y1)/(x2-x1);
    b = y1 - k * x1;
    d = abs(k * x3 - y3 + b) / sqrt((pow(k , 2) + 1));
    s = l * d / 2;
    cout << fixed << setprecision(2) << s << endl;
    return 0;
}


//思路二

#include <bits/stdc++.h>
using namespace std;

int main()
{
    double x1,y1,x2,y2,x3,y3;
    cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
    double l1,l2,l3,p,s;
    l1 = sqrt(pow((x2 - x1) , 2) + pow((y2 - y1) , 2));
    l2 = sqrt(pow((x3 - x1) , 2) + pow((y3 - y1) , 2));
    l3 = sqrt(pow((x2 - x3) , 2) + pow((y2 - y3) , 2));
    p = (l1 + l2 + l3) / 2;
    s = sqrt(p * (p - l1) * (p - l2) * (p - l3));
    cout << fixed << setprecision(2) << s << endl;
    return 0;
}

*/