设C点坐标为(M,N),则AC中点D(-1+M/2,1/2+N/2),BC中点E(3/2+M/2,5/2+N/2),因为D在y轴上,E在x轴上,所以-1+M/2=0,5/2+N/2=0,解得:M=2,N=-5,所以C(2,-5)
D(0,3),E(5/2,0)方程嘛,设DE上动点为(x,y),则有(y-3)/(x-0)=(0-3)/(5/2-0)整理得方程:y=-6x/5+3
设C点坐标为(M,N),则AC中点D(-1+M/2,1/2+N/2),BC中点E(3/2+M/2,5/2+N/2),因为D在y轴上,E在x轴上,所以-1+M/2=0,5/2+N/2=0,解得:M=2,N=-5,所以C(2,-5)
D(0,3),E(5/2,0)方程嘛,设DE上动点为(x,y),则有(y-3)/(x-0)=(0-3)/(5/2-0)整理得方程:y=-6x/5+3