P到直线y=x的距离最小时,y = lnx在P处切线与y = x平行,二者的斜率均为1
y = lnx
y' = 1/x = 1,x = 1
P(1,0)
P到直线x - y = 0的最小距离d = |1 - 0|/√(1 + 1) = √2/2
令g(x) = x - lnx
g'(x) = 1 - 1/x = 0
x = 1
即x=1时,PQ的距离最小;P(1,0),Q(1,1),距离为1