如何 通过直线外一点 绘制与该直线成一定角度的 直线？

slw7310

各位高人，请帮忙！ 如何  通过直线外一点 绘制与该直线成一定角度的 直线？    如下图，如何求出直线2啊？请不吝赐教！！谢谢。

zzyong00

你是问数学问题，还是编程代码？

slw7310

最好是 能够执行的  VBA/VB/ActiveX/API 编程代码，给个思路 已感谢不尽了！！！

slw7310

当然应该是在 autocad 环境中的VBA/VB/ActiveX/API 编程！！！

zzyong00

1. Public Sub ss()
   
2. Dim obj As AcadEntity, pt As Variant
   
3. ThisDrawing.Utility.GetEntity obj, pt, "选择直线："
   
4. Dim objL As AcadLine, objL2 As AcadLine, pt1 As Variant, dblLen As Double, dblRA As Double, pt2(2) As Double
   
5. 
   
6. If obj.ObjectName = "AcDbLine" Then
   
7.     Set objL = obj
   
8.     pt1 = ThisDrawing.Utility.GetPoint(pt, "选择点：")
   
9.     dblLen = ThisDrawing.Utility.GetDistance(pt1, "输入距离:")
   
10.     dblRA = ThisDrawing.Utility.GetAngle(pt1, "输入夹角(角度制):")
    
11.     pt2(0) = pt1(0) + dblLen * Cos(dblRA + objL.Angle)
    
12.     pt2(1) = pt1(1) + dblLen * Sin(dblRA + objL.Angle)
    
13.     Set objL2 = ThisDrawing.ModelSpace.AddLine(pt1, pt2)
    
14. End If
    
15. End Sub
    

Public Sub ss()<br /> Dim obj As AcadEntity, pt As Variant<br /> ThisDrawing.Utility.GetEntity obj, pt, "选择直线："<br /> Dim objL As AcadLine, objL2 As AcadLine, pt1 As Variant, dblLen As Double, dblRA As Double, pt2(2) As Double<br /> <br /> If obj.ObjectName = "AcDbLine" Then<br />     Set objL = obj<br />     pt1 = ThisDrawing.Utility.GetPoint(pt, "选择点：")<br />     dblLen = ThisDrawing.Utility.GetDistance(pt1, "输入距离:")<br />     dblRA = ThisDrawing.Utility.GetAngle(pt1, "输入夹角(角度制):")<br />     pt2(0) = pt1(0) + dblLen * Cos(dblRA + objL.Angle)<br />     pt2(1) = pt1(1) + dblLen * Sin(dblRA + objL.Angle)<br />     Set objL2 = ThisDrawing.ModelSpace.AddLine(pt1, pt2)<br /> End If<br /> End Sub<br />

大致是这样

slw7310

谢谢 您的回复，

aming001

算法搞通后，写程序就快了！