计算几何通用解法：

#include <bits/stdc++.h>

using u32 = unsigned;
using i64 = long long;
using u64 = unsigned long long;
using i128 = __int128;

constexpr double PI = acos(-1);

struct Point {
	double x, y;
};

// 绕原点顺时针旋转b(弧度制)角度
Point rotate(Point a, double b) {
    return {a.x * cos(b) + a.y * sin(b), -a.x * sin(b) + a.y * cos(b)};
}

void solve() {
	double a, b, c, d;
	std::cin >> a >> b >> c >> d;
	Point A = {a, b}, B = {c, d};
	// 将A作为 y 坐标小的，方便我们平移到原点
	if (b > d) {
		std::swap(A, B);
	}
	double dx = A.x, dy = A.y;
	A.x = A.y = 0;
	B.x -= dx, B.y -= dy;
	// 接下来让 B 点顺逆时针旋转60度
	Point C = rotate(B, 60 * PI / 180);
	Point D = rotate(B, -60 * PI / 180);
	if (C.x == round(C.x) && C.y == round(C.y)) {
		std::cout << C.x + dx << " " << C.y + dy << "\n";
	} else if (D.x == round(D.x) && D.y == round(D.y)) {
		std::cout << D.x + dx << " " << D.y + dy << "\n";
	} else {
		std::cout << -1 << "\n";
	}
}

int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout << std::fixed << std::setprecision(4);
	int t; 
	std::cin >> t;
	while (t --) {
		solve();
	}

	return 0;
}

其实这是一道陷阱题，我们计算等边三角形的第三个点的时候计算公式中包含$\sqrt{3}$ ，另外两个点的 $x,y$ 均为整数，只要乘以 $\sqrt{3}$ 必为无理数，不可能为整数，所以答案全输出 $-1$ 。

#include <bits/stdc++.h>

#define vi vector<int>
#define vpi vector<pair<int, int>>

using namespace std;

using ll = long long;

constexpr int N = 110;

void solve()
{
    cout << -1 << endl;
}   

signed main()
{
    int T = 1;
    cin >> T;
    while(T --) solve();
}