False
We know that the point lying on perpendicular bisector of the line segment joining two points is equidistant from these two points.
∴PA=√[−1−(0)]2+(1−2)2=√(−1)2+(−1)2=√2PB=√[3−0]2+(3−2)2=√(3)2+(1)2=√10∵PA≠PB
So, the point P does not lie on the perpendicular bisector of AB.
Alternate Method
Slope of the line segment joining the points A(-1,1) and B(3,3)
m1=3−13+1=24=12[∵m=y2−y1x2−x1]
Since, the perpendicular bisector is perpendicular to the line segment, so its slope,
m2=−1m1=−2 [by perpendicularity condition, m1m2 = - 1]
Also, the perpendicular bisector will be passing through the mid-point of the line segment joining the points A(-1,1) and B(3,3).
∴Mid-point=(−1+32,1+32)=(1,2)
[Since, mid-point of the line segment joining the points (x1,y1) and (x2,y2) is
(x1+x22,y1+y22)
Now, equation of perpendicular bisector having slope (-2) and passing through the point (1,2) is
(y−2)=(−2)(x−1)⇒y−2=−2x+2⇒2x+y=4 ..............(i)
[Since, the equation of line is (y−y1)=m(x−x1)]
If the perpendicular bisector cuts the Y-axis, then put x = 0 in Eq.(i), we get
2×0+y=4
⇒y=4
Hence, point P(0,4) is the point of intersection of Y-axis and the perpendicular bisector of line segment joining points A(-1,1) and B(3,3).