Given points A(3,−1) and B(1,1). Let P be the midpoint of AB, and Q be a point on the perpendicular bisector of AB that lies √2 units from P. Compute the midpoint P of AB : P=(
3+1
2
,
−1+1
2
)=(2,0) Compute the slope of AB : mAB=
1−(−1)
1−3
=
2
−2
=−1 Therefore, the equation of line AB is: y−1=−1(x−1)⟹x+y−2=0 The perpendicular bisector of AB must pass through P(2,0) and have slope perpendicular to -1 (i.e. slope +1 ): y−0=1(x−2)⟹y=x−2 Any point Q on this bisector satisfies y=x−2. Write Q=(x,x−2). We require the distance PQ=√2. Since P(2,0), Distance 2=(x−2)2+((x−2)−0)2=2 ⇒(x−2)2+(x−2)2=2⇒2(x−2)2=2⇒(x−2)2=1 Thus,
⇒x−2=±1⇒x=3 or x=1 If x=3, then y=3−2=1. So one solution is Q(3,1). If x=1, then y=1−2=−1. So the other solution is Q(1,−1).