设P点得坐标为(x,0)
|PA|^2=(x-2)^2+(x-1)^2
|PB|^2=(x-4)^2+(x-3)^2
(1)当PA=|PB|时, |PA|^2=|PB|^2
(x-2)^2+(x-1)^2=(x-4)^2+(x-3)^2
-6x+5=-14x+25
8x=20
x=2.5
(2)求|PA|+|PB|最小值
如图,作B点关于x轴的对称点p‘,坐标为(4,-3)
|PA|+|PB|的最小值等于|AP’|
|AP’|^2=(2-4)^2+[1-(-3)]^2=20,|AP’|=2√5
|PA|+|PB|最小值为2√5