
Mar 29th, 2012, 05:30 PM
#1
Thread Starter
Lively Member
Fourth vertex of a Rectangle
Hi
If 3 Vertices' xy coordinates are given from a rectangle, how to calculate the fourth one's xy?
The rectangle could be in any position.
Thanks.

Mar 29th, 2012, 05:39 PM
#2
Re: Fourth vertex of a Rectangle
First you need to orient the three vertices. After that it's very simple.
Consider the three vertices as three points in a triangle, A, B, and C. We need to figure out which vertex corresponds to the right angle. You can compute the angle from a point P to points Q and R as
angle(P, Q, R) = Acos[(QP) dot (RP) / [QP * RP]]
where Acos is the inverse cosine, (QP) etc. is vector subtraction (subtract componentwise), QP etc. is the magnitude of a vector (sum the squares of the components and take the square rootthis is the content of the Pythagorean theorem), and (QP) dot (RP) is the dot product (multiply the x components together, then the y components, etc. and add the results).
Compute angle(A, B, C), angle(B, C, A), and angle(C, A, B). One of these should be very, very close to 90 degrees (pi/2 radians). If it's the first one, the vertex is at A; the second gives a vertex at B, and the third a vertex at C.
Now suppose the vertex is at P and the other two points are Q and R. The fourth point on the rectangle is at Q + (RP).
The time you enjoy wasting is not wasted time.
Bertrand Russell
< Remember to rate posts you find helpful.

Mar 29th, 2012, 06:23 PM
#3
Thread Starter
Lively Member
Re: Fourth vertex of a Rectangle
Thanks, really good idea to calculate 90 angle. But write in VB takes time.

Mar 29th, 2012, 07:46 PM
#4
Re: Fourth vertex of a Rectangle
If you need help with coding my idea, you can ask in the appropriate forum (the .NET or VB6 ones depending on your language). If you need a formula expanded or explained you can ask here in the maths forum.
The time you enjoy wasting is not wasted time.
Bertrand Russell
< Remember to rate posts you find helpful.

Mar 29th, 2012, 08:40 PM
#5
Thread Starter
Lively Member
Re: Fourth vertex of a Rectangle
Ye, I will try to code first.
Thanks

Mar 30th, 2012, 10:15 AM
#6
Thread Starter
Lively Member
Re: Fourth vertex of a Rectangle
Code:
Partial Public Class Window2
Public Sub New()
' This call is required by the Windows Form Designer.
InitializeComponent()
' Add any initialization after the InitializeComponent() call.
End Sub
Private Sub msg()
Dim Point1 As Point = New Point(1, 1)
Dim Point2 As Point = New Point(4, 1)
Dim Point3 As Point = New Point(1, 5)
Dim PointWithBiggestAngle As Integer = GetPointOnBiggestAngle(Point1, Point2, Point3)
Dim P, Q, R As Point
Select Case PointWithBiggestAngle
Case 1
P = Point1
Q = Point2
R = Point3
Case 2
P = Point2
Q = Point1
R = Point3
Case 3
P = Point3
Q = Point2
R = Point1
End Select
Dim Point4X As Long = Q.X + (R.X  P.X)
Dim Point4Y As Long = Q.Y + (R.Y  P.Y)
Debug.WriteLine(Point4X & " " & Point4Y)
End Sub
Private Sub Window2_Loaded(ByVal sender As Object, ByVal e As System.Windows.RoutedEventArgs) Handles Me.Loaded
msg()
End Sub
Private Function GetPointOnBiggestAngle(ByVal P As Point, ByVal Q As Point, ByVal R As Point) As Integer
Dim AngleOfP As Double = GetAngle(P, Q, R)
Dim AngleOfQ As Double = GetAngle(Q, P, R)
Dim AngleOfR As Double = GetAngle(R, Q, P)
If AngleOfP > AngleOfQ AndAlso AngleOfP > AngleOfR Then
Return 1
ElseIf AngleOfQ > AngleOfP AndAlso AngleOfQ > AngleOfR Then
Return 2
ElseIf AngleOfR > AngleOfP AndAlso AngleOfR > AngleOfQ Then
Return 3
Else
Return 0
End If
End Function
Private Function GetAngle(ByVal anglePoint As Point, ByVal Q As Point, ByVal R As Point) As Double
Dim a As Double = Math.Sqrt((Q.X  R.X) ^ 2 + (Q.Y  R.Y) ^ 2) 'anglePoint's opposite side length
Dim b As Double = Math.Sqrt((anglePoint.X  R.X) ^ 2 + (anglePoint.Y  R.Y) ^ 2) ' Q's opposite side length
Dim c As Double = Math.Sqrt((Q.X  anglePoint.X) ^ 2 + (Q.Y  anglePoint.Y) ^ 2)
Return Math.Acos((b ^ 2 + c ^ 2  a ^ 2) / (2 * b * c))
End Function
End Class
Last edited by sharethl; Mar 30th, 2012 at 10:19 AM.
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