1. Drop the perpendicular from M to the x -axis, and the perpendicular from N to the y-axis.
Let these perpendiculars intersection be point A, with coordinates (2, 4).
[clearly A will have x coordinate equal to the x coordinate of M, and y coordinate equal to the y coordinate of N]
2. Let P be the point on NM such that |NP|/|PM|=1/3
3. Drop the perpendiculars from point P to AN and AM. Let the foot of these perpendiculars be points P' and P'' respectively.
4. By Thales theorem: |NP'|/|P'A|=1/3 and |AP''|/|P''M|=1/3
so let |NP'|=b, |P'A|=3b, |AP''|=c, |P''M|=3c, as shown in the figure
5. |AN|=14-2=12.
|AN|=4b so b=12/4=3
the x coordinate of P'=2+3b=2+9=11
the x coordinate of P= the x coordinate of P'=11
similarly |AM|=8-4=4
|AM|=4c so c=1
the y coordinate of P'' is 4+c=4+1=5
the y coordinate of P = the y coordinate of P''=5
Answer: (11, 5)
