How can I position the window's position on startup to the right side of the user's screen?

Description

You can use Screen from System.Windows.Forms.

So add reference to the System.Windows.Forms.dll and System.Drawing.dll. Then change the Left and Height property in the MainWindow_Loaded method.

Sample

public MainWindow(){    InitializeComponent();    this.Loaded += new RoutedEventHandler(MainWindow_Loaded);}void MainWindow_Loaded(object sender, RoutedEventArgs e){    this.Left = System.Windows.Forms.Screen.PrimaryScreen.WorkingArea.Right - this.Width;    this.Top = 0;    this.Height = System.Windows.Forms.Screen.PrimaryScreen.WorkingArea.Height;}

More Information

• MSDN - Screen Class

You can do this without referencing win forms assemblies by using SystemParameters. In the code behind for your window XAML:

MainWindow() {    InitializeComponents();    this.Loaded += new RoutedEventHandler(      delegate(object sender, RoutedEventArgs args) {        Width = 300;        Left = SystemParameters.VirtualScreenLeft;        Height = SystemParameters.VirtualScreenHeight;    }}

SystemParameters documentation

in your xaml :

WindowStartupLocation="Manual" 

in the constructor :

 Left = System.Windows.SystemParameters.PrimaryScreenWidth - Width Top=0