Area of a Parallelogram

http://acm.hust.edu.cn:8080/judge/contest/view.action?cid=8086#problem/H

 

Description

A parallelogram is a quadrilateral with two pairs of parallel sides. See the picture below:

 

Fig: a parallelogram

Now you are given the co ordinates of A, B and C, you have to find the coordinates of D and the area of the parallelogram. The orientation of ABCD should be same as in the picture.

Input

Input starts with an

 integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing six integers A_x, A_y, B_x, B_y, C_x, C_y where (A_x, A_y) denotes the coordinate of A, (B_x, B_y) denotes the coordinate of B and (C_x, C_y) denotes the coordinate of C. Value of any coordinate lies in the range [-1000, 1000]. And you can assume that A, B and C will not be collinear.

Output

For each case, print the case number and three integers where the first two should be the coordinate of D and the third one should be the area of the parallelogram.

Sample Input

3

0 0 10 0 10 10

0 0 10 0 10 -20

-12 -10 21 21 1 40

Sample Output

Case 1: 0 10 100

Case 2: 0 -20 200

Case 3: -32 9 1247

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std ;

typedef struct Coord{
    int x , y ;
} Coord ;

Coord c[4] ;

int CalcuArea()
{
    double lab = sqrt( (c[0].x-c[1].x)*(c[0].x-c[1].x)  + (c[0].y-c[1].y)*(c[0].y-c[1].y) ) ;
    double lad = sqrt( (c[0].x-c[3].x)*(c[0].x-c[3].x)  + (c[0].y-c[3].y)*(c[0].y-c[3].y) ) ;
    double d = fabs( (c[3].x-c[0].x)*(c[1].x-c[0].x) + (c[3].y-c[0].y)*(c[1].y-c[0].y) ) / lab ;
    double h = sqrt( lad*lad - d*d ) ;
    int ar = h * lab+0.5 ;
    return ar ;
}

int main()
{
    int T , i , ar , n ;
    scanf("%d", &T ) ;
    n = T ;
    while(T -- )
    {
        for(i = 0 ; i < 3 ; i ++ ) scanf("%d %d" , &c[i].x , &c[i].y ) ;
        c[3].x = c[2].x - c[1].x + c[0].x ;
        c[3].y = c[2].y - c[1].y + c[0].y ;
        ar = CalcuArea() ;
        printf("Case %d: %d %d %d\n", n - T , c[3].x , c[3].y , ar) ;
    }
    return 0 ;
} 

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std ;

typedef struct Coord{
    int x , y ;
} Coord ;

Coord c[4] ;

int CalcuArea()
{
    int lab =abs((c[3].x-c[0].x)*(c[1].y-c[0].y)-(c[3].y-c[0].y)*(c[1].x-c[0].x));
    return lab;
}

int main()
{
    int T , i , ar , n ;
    scanf("%d", &T ) ;
    n = T ;
    while(T -- )
    {
        for(i = 0 ; i < 3 ; i ++ ) scanf("%d %d" , &c[i].x , &c[i].y ) ;
        c[3].x = c[2].x - c[1].x + c[0].x ;
        c[3].y = c[2].y - c[1].y + c[0].y ;
        ar = CalcuArea() ;
        printf("Case %d: %d %d %d\n", n - T , c[3].x , c[3].y , ar) ;
    }
    return 0 ;
}