#include <iostream>
using namespace std;
//下面就是要寻找的二次函数型数组，目标是找到其中的最大值
int a[10] = { 0, 3, 25, 21, 20, 14, 12, 7, 2, 1 };
int main()
{
	int z = 0, y = 9;
	while (z <= y)
	{
    //三分，就是将区间尽量平均分成三份
		int mid1 = z + (y - z) / 3;
		int mid2 = y - (y - z) / 3;
    //判断两者中的较大值，那么较小值的一方的元素都不可能是最大值，直接舍去
		if (a[mid1] >= a[mid2]) y = mid2 - 1;
		else z = mid1 + 1;
	}
  //最终结果位置与二分类似
	cout << a[z] << endl;
	return 0;
}

double f(double x) {
    // 示例函数，此处替换为实际函数
    return -x*x + 2*x + 3; // 极大值在x=1处
}
// 精度控制
const double eps = 1e-9;
while (r - l > eps)
{
    double m1 = l + (r - l) / 3;
    double m2 = r - (r - l) / 3;
    if (f(m1) < f(m2))
    {
        // 求极大值，若求极小值则改为>,极值在[m1, r]
        l = m1;
    } else
    {
      // 极值在[l, m2]
        r = m2;
    }
}
// 返回近似极值点
return (l + r) / 2;

int T = 100; // 迭代次数，确保精度
double f(double x)
{
    return -x*x + 2*x + 3; // 示例函数，极大值在x=1
}
double ternary_search_fixed_loop(double l, double r, int T) {
while (T--)
{
    double m1 = l + (r - l) / 3;
    double m2 = r - (r - l) / 3;
    if (f(m1) < f(m2))
    { // 求极大值
        l = m1; // 极值在右区间
    } else
    {
        r = m2; // 极值在左区间
    }
}
return (l + r) / 2;
}

#include <iostream>
#include <cmath>
using namespace std;
const int N = 5005;
struct ty
{
	double x, y;
}a, b, c, d, e[N + 5], f[N + 5];
//t1是当前点到A点所需的时间，t2是当前点到D点所需的时间
double t1[N + 5], t2[N + 5];
double p, q, r;
//计算距离的函数
double dis(ty a, ty b)
{
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
//第二个三分，用于嵌套在第一个三分内部，返回总时间最小的情况
double sol(int mid)
{
	int z2 = 0, y2 = N;
	while (z2 <= y2)
	{
		int mid1 = z2 + (y2 - z2) / 3;
		int mid2 = y2 - (y2 - z2) / 3;
		if ((t1[mid] + t2[mid1] + (dis(e[mid], f[mid1]) / r)) <= (t1[mid] + t2[mid2] + (dis(e[mid], f[mid2]) / r)))
			y2 = mid2 - 1;
		else z2 = mid1 + 1;
	}
	return (t1[mid] + t2[z2] + (dis(e[mid], f[z2]) / r));
}
int main()
{
	cin >> a.x >> a.y >> b.x >> b.y;
	cin >> c.x >> c.y >> d.x >> d.y;
	cin >> p >> q >> r;
  //关键步骤！！！用于将实数三分转换为整数型三分，最后分别存放于数组 e 和 f 中。
	double dx = (b.x - a.x) / N, dy = (b.y - a.y) / N;
	for (int i = 0; i <= N; i++)
	{
		e[i].x = a.x + dx * i, e[i].y = a.y + dy * i;
		t1[i] = dis(a, e[i]) / p;
	}
	dx = (d.x - c.x) / N, dy = (d.y - c.y) / N;
	for (int i = 0; i <= N; i++)
	{
		f[i].x = c.x + dx * i, f[i].y = c.y + dy * i;
		t2[i] = dis(d, f[i]) / q;
	}
  //外层三分循环，枚举AB段的情况，再结合sol函数，最终求出时间最小的情况！
	int z1 = 0, y1 = N;
	while (z1 <= y1)
	{
		int mid1 = z1 + (y1 - z1) / 3;
		int mid2 = y1 - (y1 - z1) / 3;
		if (sol(mid1) <= sol(mid2)) y1 = mid2 - 1;
		else z1 = mid1 + 1;
	}
  //按照题目要求精度输出即可
	printf("%.2f\n", sol(z1));
	return 0;
}