Answer: Their equations have different y-intercept but the same slope
Step-by-step explanation:
Since, the slope of the line passes through the points (2002,p) and (2011,q) is,

Similarly, the slope of the line asses through the points (2,p) and (11,q) is,

Since, 
Hence, both line have the same slope.
Now, the equation of the line one having slope
and passes through the point (2002,p) is,

Put x = 0 in the above equation,
We get, 
The y-intercept of the line one is 
Also, the equation of second line having slope
and passes through the point (2,p)

Put x = 0 in the above equation,
We get, 
The y-intercept of the line one is 
Thus, both line have the different y-intercepts.
⇒ Third option is correct.