C# 通过坐标计算角度

例如在panle中，鼠标在其中任意位置点击2次，获得2组Point坐标数据，通过此2点画条直线，计算此直线相对于水平面(也就是x轴)的角度,只求能告诉我如何计算角度，公式是什么，直角坐标系的坐标和C#的坐标系不太一样，这块怎么处理？在线等，谢谢

double angleOfLine = Math.Atan2((Y2 - Y1), (X2 - X2)) * 180 / Math.PI

try { //参数赋值 double ss1 = 0.00000001, ss2 = 1000000,we=7.2921151467*ss1*1000; double pi = 3.1415926535; string[] str = textbox2.text.trim().split(' '); double[,] a = new double[7, 4]; int k = 0; for (int i = 0; i < 7; i++) { for (int j = 0; j < 4; j++) { a[i, j] = convert.todouble(str[k]); k++; } } //规划时刻计算 int[] dinmth = new int[12]; double year, month, day, hour, minute, second, dayofy, ttlday, dayofw, gpstime, tk; int weekno; dinmth[0] = 31; dinmth[1] = 28; dinmth[2] = 31; dinmth[3] = 30; dinmth[4] = 31; dinmth[5] = 30; dinmth[6] = 31; dinmth[7] = 31; dinmth[8] = 30; dinmth[9] = 31; dinmth[10] = 30; dinmth[11] = 31; year = convert.todouble(textbox21.text); month = convert.todouble(textbox22.text); day = convert.todouble(textbox23.text); hour = convert.todouble(textbox24.text); minute = convert.todouble(textbox25.text); second = convert.todouble(textbox26.text); if (month == 1) { dayofy = day; } else { dayofy = 0; for (int i = 0; i < 12; i++) { dayofy += dinmth[i]; if (i == 1) { if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) dayofy += 1; } } dayofy += day; } ttlday = 360; for (int yr = 1981; yr < year; yr++) { ttlday += 365; if (yr % 4 == 0 && yr % 100 != 0 || yr % 400 == 0) { ttlday += 1; } } ttlday += dayofy; weekno = (int)(ttlday / 7); dayofw = ttlday - 7 * weekno; gpstime = hour * 3600 + minute * 60 + second + dayofw * 86400; tk = gpstime - a[3, 0] * ss2; if (tk > 302400) { tk -= 604800; } else if(tk<-302400) { tk += 602400; } app.value1 = tk.tostring("e2"); //平均角速度计算 double n0, n; n0 = (math.sqrt(3986004.418) / math.pow(a[2, 3], 3)) * ss1; n = n0 + a[1, 2] * ss1; app.value2 = n.tostring("e2"); //平近点角计算 double m0, mk; m0 = a[1, 3]; mk = m0 * 10 + n * tk; if (math.abs(mk) > 2 * pi) { int w1 = (int)(mk / (2 * pi)); mk = mk - 2 * pi * w1; } app.value3 = mk.tostring("e2"); //偏近点角计算 double ek, e0; ek = mk; do { e0 = ek; ek = mk + a[2, 1] * 0.1 * math.sin(e0); } while (math.abs(ek - e0) > 0.00000001); app.value4 = ek.tostring("e2"); //真近点角 double angle1, angle2, vk; angle1 = math.sqrt(1 - math.pow(a[2, 1] * 0.1, 2)) * math.sin(ek); angle2 = math.cos(ek) - a[2, 1] * 0.1; vk = math.atan(math.abs(angle1 / angle2)); if (angle1 >= 0 && angle2 <= 0) { vk = pi - vk; } else if (angle1 <= 0 && angle2 <=0) { vk = pi + vk; } else { vk = 2 * pi - vk; } app.value5 = vk.tostring("e"); //升交角距 double qk; qk = vk + a[4, 2] * 10; if (math.abs(qk) > 2 * pi) { int w2 = (int)(qk / (2 * pi)); qk = qk - 2 * pi * w2; } app.value6 = qk.tostring("e"); //升交角距改正,轨道向径改正，轨道倾角改正计算 double fu, fr, fi; fu = a[2, 0] * ss1 * 1000 * math.cos(2 * qk) + a[2, 2] * ss1 * 1000 * math.sin(2 * qk); fr = a[4, 1] * 1000 * math.cos(2 * qk) + a[1, 1] * 100 * math.sin(2 * qk); fi = a[3, 1] * ss1 * 100 * math.cos(2 * qk) + a[3, 3] * ss1 * 10 * math.sin(2 * qk); app.value7 = fu.tostring("e"); app.value8 = fr.tostring("e"); app.value9 = fi.tostring("e"); //改正后升交角距、轨道向径、轨道倾角计算 double uk, rk, ik; uk = qk + fu; rk = math.pow(a[2, 3] * ss2 * 0.01, 2) * (1 - a[2, 1] * 0.1 * math.cos(ek)) + fr; ik = a[4, 0] + fi + a[5, 0] * ss1 * 0.1 * tk; app.value10 = uk.tostring("e"); app.value11 = rk.tostring("e"); app.value12 = ik.tostring("e"); //升交点轨道直角坐标系坐标计算 double xk, yk; xk = rk * math.cos(uk); yk = rk * math.sin(uk); app.value13 = xk.tostring("e"); app.value14 = yk.tostring("e"); //升交点经度计算 double n; int w3; n = a[3, 2] * 10 + (a[4, 3] * ss1 - we) * tk - we * a[3, 0] * ss2; if (math.abs(n) > 2 * pi) { w3 =(int)(n / (2 * pi)); n = n - 2 * pi * w3; } app.value15 =n.tostring("e"); //地固坐标系下卫星空间直角坐标计算,ik为ik double x, y, z; x = xk * math.cos(n) - yk * math.cos(ik) * math.sin(n); y = xk * math.sin(n) + yk * math.cos(ik) * math.cos(n); z = yk * math.sin(ik); app.value16 = x.tostring("e"); app.value17 = y.tostring("e"); app.value18 = z.tostring("e"); new form2().showdialog(); } catch { messagebox.show("请输入卫星时刻或又一次加载卫星参数导致出错"); return; }