You will see this if you draw the phasor diagrams.
Let's just model your transmission line as an inductor with an impedence of Xl*i.
Zt = Transmission line impedence = Xl*i= 1<90 deg pu. Completely reactive transmission line.
Vs = source voltage. = 1<0 deg pu for this example.
Il = load current. = 0.1<-90 deg pu for this example. A completely reactive load. Current lagging voltage by full 90 deg at 0.1 pu.
Vl = Voltage at load = Vs - Il*Zt = 1<0 - (0.1<-90)*(1<90) = 0.9 < 0 deg pu.
If you draw you out that phasor diagram for Vl = Vs-Il*Zt, what you will see is that the lagging current puts the voltage drop more inline with the source voltage resulting in a greater magnitude drop. If the load had been purely resistive, the voltage drop across the transmission line would have been 90 degree out of phase with the source voltage and the voltage drop would have been magnitudely less. Let's say that load was capacitive, like a capacitor bank, the voltage drop would be 180 degrees out of phase with the source and the voltage would actually be higher at the cap bank than at the source. The voltage drop will always be the greatest when the angle of the current is equal and opposite of the angle of the transmission line impedance. It just happens to be that transmission lines generally are very reactive. This is also why the voltage drop across a transformer depends on the power factor of the load.
To further answer your question of real and imaginary power flows, the angle differences between buses tends to be small. This allows the power flow equations to be decoupled for iterative power flow solution methods. When you decouple the equations, vars flow from high voltage to low voltage (on a pu basis) and real power flows based on voltage angle differences between buses. If you ever use do transmission studies with PSS/, Powerworld, or the like, it becomes very obvious that capacitor banks hardly impact real power flows and you need real power generation or load to change real power flows, excluding outages.
Look under 'Power-flow problem formulation' to see how the equations can be decoupled
