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

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

Sample Input	Sample Output
Click to copy 3 0 0 10 0 10 10 0 0 10 0 10 -20 -12 -10 21 21 1 40	Click to copy Case 1: 0 10 100 Case 2: 0 -20 200 Case 3: -32 9 1247

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LOJ 1305 - Area of 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.

Solution

Since, it's a parallelogram,

    AB = CD
=>  Bx - Ax = Cx - Dx
=>  Dx = Ax + Cx - Bx

Similarly, D_y = A_y + C_y - B_y.

We know that the area of a Parallelogram,

Simply A = 1/2 * (((A_xB_y)+(B_xC_y)+(C_xD_y)+(D_xA_y))-((A_yB_x)+(B_yC_x)+(C_yD_x)+(D_yA_x)))

We have to assign A = A * -1 in case A comes in negative.

If we now just simply plugin the values, we have all the answer now.

The above implementation is accepted.

Solution in C

#include <stdio.h>

int main()
{
    int t, ax, ay, bx, by, cx, cy, dx, dy, a;
    scanf("%d", &t);
    for (int i = 1; i <= t; i++)
    {
        scanf("%d %d %d %d %d %d", &ax, &ay, &bx, &by, &cx, &cy);
        dx = ax + cx - bx;
        dy = ay + cy - by;
        a = 0.5 * (((ax * by) + (bx * cy) + (cx * dy) + (dx * ay)) - ((ay * bx) + (by * cx) + (cy * dx) + (dy * ax)));
        if (a < 0)
            a *= -1;
        printf("Case %d: %d %d %d\n", i, dx, dy, a);
    }
    return 0;
}

Tutorial source