试题分析:根据A(m1,n1),B(m2,n2)在直线y=kx+b上,,可得出n1,n2的值,再得出n1+n2=k(m1+m2)+2b,故可得出k+1=,再根据b>2可知0<<1,故可得出0<k+1<1,再由m1<m2即可得出结论. 试题解析:∵A(m1,n1),B(m2,n2)在直线y=kx+b上, ∴n1=km1+b,n2=km2+b. ∴n1+n2=k(m1+m2)+2b. ∴kb+4=3kb+2b. ∴k+1=. ∵b>2, ∴0<<1. ∴0<k+1<1. ∴-1<k<0. ∵m1<m2, ∴n2<n1. |