13. Well I really hope you can show that the midpoint of (0,0) and (x, 4x+3) has the coordinates given.
The only thing that should confuse you after that is that the parameter is x.
To avoid confusion, call it t instead: ie. x=t/2, y=(4t+3)/2
Then eliminate the parameter.
14. Changing the subject gives you y=(2/3) (x+3).
So P has coordinates [ x, (2/3) (x+3)].
Then its almost exactly the same as Q13, except you use the division in a ratio formula instead of midpoint.
