Hi Nick,
When you have plate curves, you have all !
In attached one (It's a Russian 6N6pi)
Pick a first point on a grid line, f.e. the red one where Vg is -4V, Vp is +155V and Ip is 30mA
Move down vertically (constant plate voltage) till you cross another grid line (the blue dot).
Here Vg is -6V and Ip is 12 mA.
That means that plate current change is (30 -12) 18mA for Vg change of 2 Volts.
Thus the slope is (18 / 2) 9mA per Volt, the same that a transcondustance of 9000 if expressed in mmoh.
Then, move left horizontally (constant plate current) till you meet the original grid line (green dot).
The plate voltage is now 120V for a 2V grid voltage difference, thus the Mu, µ or amplification coefficient is (155-120) / 2 = 17.5
The Rp is figured by the slope of a straight line joining the red and the grid dot.
This is the same as dividing the Mu by the slope (in Ampere per Volt), here 17.5/.009 = 1K94.
<edit>
Oopps, misread the plate voltage !
</edit>
Yves.