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

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second
#define FILL(a, b) memset(a, b, sizeof(a))

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;

struct point
{
	int x, y;
	
	point() {}
	point(int a, int b): x(a), y(b) {}
	
	LL dis(point p)
	{
		return LL(abs(x - p.x)) + abs(y - p.y);
	}
};

point A, B, C;
double a, b;
int p = 0;
double ans[100447];

double f(point X)
{
	double d1 = A.dis(X) / a;
	double d2 = B.dis(X) / b;
	double d3 = C.dis(X) / b;
	return max(d1, d2) + d3;
}

double ter(double x)
{
	int ly = min({A.y, B.y, C.y}), ry = max({A.y, B.y, C.y});
	double res = 2e9;
	while (ly + 4 < ry)
	{
		int d = (ry - ly) / 2;
		int m1 = ly + d;
		int m2 = ly + d + 1;
		double f1 = f(point(x, m1));
		double f2 = f(point(x, m2));
		res = min({res, f1, f2});
		
		if (f1 < f2)
			ry = m2;
		else
			ly = m1;
	}
	FOR (i, ly, ry + 1)
		res = min(res, f(point(x, i)));
	ans[p] = min(ans[p], res);
	return res;
}


void solve()
{
	scanf("%lf%lf", &a, &b);
	scanf("%d%d", &A.x, &A.y);
	scanf("%d%d", &B.x, &B.y);
	scanf("%d%d", &C.x, &C.y);
	ans[p] = A.dis(C) / a;
	int lx = min({A.x, B.x, C.x}), rx = max({A.x, B.x, C.x});
	while (lx + 4 < rx)
	{
		int d = (rx - lx) / 2;
		int m1 = lx + d;
		int m2 = lx + d + 1;
		if (ter(m1) < ter(m2))
			rx = m2;
		else
			lx = m1;
	}
	FOR (i, lx, rx + 1)
		ter(i);
	p++;
}

int main()
{
	//ios::sync_with_stdio(false);
	//cin.tie(0);
	
	int t;
	scanf("%d", &t);
	FOR (tt, 0, t)
	{
		solve();
	}
	FOR (i, 0, t) printf("%f\n", ans[i]);
	
	return 0;
}