设y-2=k(x+1)
则与两直线交点横坐标分别为-(3k+9)/(3k+4),(-k+1)/(k+1)
∴-(3k+9)/(3k+4)+(-k+1)/(k+1)=-2
解得k=-3
所以所求直线为y-2=-3x-3
即3x+y+1=0