Question

A(2,1) B(5,2) 直线L：(K+2)x-(k+1)y+2k-1=0与线段AB有公共交点 求直线斜率的变化范围

Answer 1

解由线性规划理论可知

把A(2,1)代入(K+2)x-(k+1)y+2k-1中得(K+2)*2-(k+1)*1+2k-1=3k+2

把A(5,2)代入(K+2)x-(k+1)y+2k-1中得(K+2)*5-(k+1)*2+2k-1=5k+7

则3k+2与5k+7的关系为

（3k+2）（5k+7）≤0

即-5/7≤k≤-2/3