Y+1=0和X+Y+1=0的交点O(0,-1)
A(-1,-4),B(b.-1)
k(AB)=3/(1+b),k(BC)=c/(b-c),k(AC)=(3-c)/(1+c)
角B平分线方程:Y+1=0
k(AB)=-k(BC)
3/(1+b)=-c/(b-c)
3b+bc-2c=0.(1)
角C平分线方程:X+Y+1=0
[-1-k(AC)]/[1-k(AC)]=[k(BC)+1]/[1-k(BC)]
b+bc-4c=0.(2)
b=5,c=-5
B(5,-1),C(-5,4)
k(AB)=0.5
k(bc)=-5/(5+5)=-1/2
AB:x-2y-7=0
BC:y+1=-1/2(x-5)
y=-1/2x+3/2