Question 268949
Took a while, but I have a possible answer.
Let 
(i) M = (0,0) 
and 
(ii) P = (1,5).
This makes the slope of MP = 5.
Now we need a coordinate for 
(iii) N = (x,y) 
such that
(iv) m(M,N) = 1/2
and
(v) m(N,P) = 2/3
----
from (i), (iii), and (iv) 
(vi) m(M,N) = (y-0)/(x-0) =  1/2
(vii) x = 2y
----
from (ii), (iii), and (v) 
(viii) m(N,P) = (y-5)/(x-1) =  2/3
and by substitution of (vii) into (viii) we get
(ix) y-5)/(2y-1) =  2/3
cross multiply to get
(x) 3y -15 = 4y - 2
subtract 3y and add 2 to get
(xi) y = -13
This means that x = -26 and our point N is
N = (-26, -13)
---
So in total we get
M = (0,0) 
and 
P = (1,5)
and
N = (-26,-13)
m(M,P) = 5/1 = 5
m(M,N) = 13/26 = 1/2
m(N,P) = 18/27 = 2/3