A.MATHS問題....唔該!!

let the point be (x1,y1)

i.e. (x1-2)^2+(y1-2)^2 = (x1-3)^2+(y1+1)^2

since (x1,y1) is on the line x+y=1
we will have
x1+y1=1 => y1=1-x1

hence,
(x1-2)^2+[(1-x1)-2]^2 = (x1-3)^2+[(1-x1)+1]^2
(x1-2)^2+(-x1-1)^2 = (x1-3)^2+ (2-x1)^2
x1^2-4x1+4+x1^2+2x1+1=x1^2-6X1+9+x1^2-4x1+4
8x1=8
x1=1
y1=1-(1)=0

i.e. the point is (1,0)

the point is (1,0)