Answer:
we have
(y-y1)/(x-x1)=(y2-y1)/(x2-x1)
(y-1)/(x--2)=(-1-1)/(1--2)
(y-1)/(x+2)=(-2)/3
3(y-1)=(-2(x+2))
3y-3=-2x-4
2x+3y-3+4=0
2x+3y+1=0 is a required equation
4.7m questions
6.1m answers