Question

已知点A(-2,5)和B(2,4),在直线l:2x-2y+1=0上求一点P,使|PA|+|PB|最小,并求出这个最小值.

Answer 1

设P(x,y)

又2x-2y+1=0

y=x+1/2

所以：|PA|+|PB|=√{[(x+2)^2+(y-5)^2]+[(x-2)^2+(y-4)^2]}

=√{[(x+2)^2+(x-9/2)^2]+[(x-2)^2+(x-7/2)^2]}

=√(4x^2+4x-9x-4x-7x+4+81/4+4+49/4)

=√(4x^2-16x+8+65/2)

=√(4x^2-16x+16-16+8+65/2)

==√[4(x-2)^2+33/2]>=√66/2(当且仅当x=2,y=5/2时取等号)

|PA|+|PB|最小值为√66/2