How to use IV characteristics of a diode in calculations
The graph give you a more accurate answer that assuming 0.7 which is only an approximation.  The voltage and resistor are already in thevenin equivalent form.  on the graph this is a line as the voltage goes from 0 to 50.  It would be easy to plot the line if the graph when out to 50V because then out just use the end points 0.1A and 50V.  Since this ins impractical on this graph, use the equation I=-V/500+0.1.  If you do this at V=1.2 I=0.0976A.  Plot this point and draw the line to 0V,0.1A.  Where this line crosses the diode curve is the actual voltage for the diode when you have 50V and 500Ohms.