[vb.net] How to open child forms positioned within MDI parent in VB.NET?

Try the code below and.....

1 - change the name of the MENU as in my sample the menuitem was called 'Form7ToolStripMenuItem_Click'

2 - make SURE to paste it into an MDIFORM and not just a basic FORM

Then let me know if the CHILD form still shows OUTSIDE the parent form

Private Sub Form7ToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Form7ToolStripMenuItem.Click

    Dim NewForm As System.Windows.Forms.Form
    NewForm = New System.Windows.Forms.Form
    'USE THE NEXT LINE - to add an existing CUSTOM form you already have
    'NewForm = Form7                         
    NewForm.Width = 400
    NewForm.Height = 250
    NewForm.MdiParent = Me
    NewForm.Text = "CAPTION"
    NewForm.Show()
    DockChildForm(NewForm, "left")          'dock left
    'DockChildForm(NewForm, "right")         'dock right
    'DockChildForm(NewForm, "top")           'dock top
    'DockChildForm(NewForm, "bottom")        'doc bottom
    'DockChildForm(NewForm, "full")          'fill the client area (maximise the child INSIDE the parent)
    'DockChildForm(NewForm, "Anything-Else") 'center the form

End Sub

Private Sub DockChildForm(ByRef Form2Dock As Form, ByVal Position As String)

    Dim XYpoint As Point
    Select Case Position
        Case "left"
            Form2Dock.Dock = DockStyle.Left
        Case "top"
            Form2Dock.Dock = DockStyle.Top
        Case "right"
            Form2Dock.Dock = DockStyle.Right
        Case "bottom"
            Form2Dock.Dock = DockStyle.Bottom
        Case "full"
            Form2Dock.Dock = DockStyle.Fill
        Case Else
            XYpoint = New Point
            XYpoint.X = ((Me.ClientSize.Width - Form2Dock.Width) / 2)
            XYpoint.Y = ((Me.ClientSize.Height - Form2Dock.Height) / 2)
            Form2Dock.Location = XYpoint
    End Select
End Sub