三角形ABC顶点的坐标分别为A(0,0),B(4,8),C(6,-4),点M分AB的比为3,N是边AC上一点,

S△AMN=AM*AN*sinBAC/2
S△ABC=AB*AC*sinBAC/2
S△AMN:S△ABC=(AM/AB)*(AN/AC)=1/2
M分AB所成比为3
AM/AB=3/4
∴AN/AC=2/3
N分AC所成比为2,A(0,0),C(6,-4)
N[(0+2*6)/(1+2),(0-2*4)/(1+2)]
即N(4,-8/3)
直线MN的方程就好求了