证明：对任意实数m,直线(m+2)x-(m+1)y-2(3+2m)=0与P(-2,2)的距离d恒小于4√2
问题描述：
证明：对任意实数m,直线(m+2)x-(m+1)y-2(3+2m)=0与P(-2,2)的距离d恒小于4√2
答
距离d=|-2(2+m)-2(1+m)-2(3+2m)|/√[(2+m)^2+(1+m)^2] =|-4-2m-2-2m-6-4m|/√(4+4m+m^2+1+2m+m^2) =|8m+12|/√(2m^2+6k+5) 令k=d^2=(64m^2+192m+144)/(2m^2+6m+5) 64m^2+192m+144=2km^2+6mk+5k (64-2k)m^2+(192-6k)m...