Don't get complicate. Apply ohm's law, Which states I = V/R. if ur Vcc = +5V. current through the string, I = 5/11KΩ => 0.4545mA .Take upper resistor as R1 and lower resistor as R2
1. If you moving the adjustable end of POT towards +Vcc(i.e., fully turned to one side), then it becomes,
voltage drop across R1, Vr1 = I * R1 => 0.4545mA * 10K => 4.54V. and drop across R2, Vr2 = 0.45V.
+5V------^^^^^^^^--------^^^^^--------gnd
---------10K---------------------1K-----------
------- R1------------------------ R2 -----------
2. If POT is exactly at center, then it becomes, take a rough sheet and draw, split the POT into two, as
+5V------^^^^----^^^^--------^^^^^--------gnd
------------5K--------5K-------------1K-----------
------- --- R1----------------------- R2 -----------
Vr1 = 0.4545mA * 5K => 2.2727V
As in your figure, it shows you are not measuring w.r.t ground. But i think while you simulating, all the voltages are w.r.t ground.
3. When POT is moved fully to another side, drop is zero .
+5V------^^^^^^^^--------^^^^^--------gnd
-------------------10K---------------1K-----------
--------------- R1 ----------------- R2 -----------
Best wishes