Point P(0,2) is the point of intersection of Y-axis and perpendicular bisector of line segment joining the points A(-1,1) and B(3,3).

Explanation

We know that , the points lies on perpendicular bisector of the line segment joining the two points is equidistant from these two points.
`:. PA =sqrt([-4-(4)]^(2)+(6-2)^(2))`
`=sqrt((0)^(2)+(4)^(2))=4`
PB `=sqrt([-4-4]^(2)+(-6-2)^(2))sqrt((0)^(2)+(-8)^(2))=8` `:' PAnePB`
So , the point P does not lie on the perpendicular bisctor of AB.
Alternate Method
Slope of the line segment joining the points A(-1,1) and B(3,3), `m_(1)=(3-1)/(3+1)=(2)/(4)=(1)/(2)[:' m=(y_(2)-y_(1))/(x_(2)-x_(1))]`
Since , the perpendicular bisector is perpendicular to the line segment , so its slope,
`m_(2)=-(1)/((y_(2)))=-2` [by perpendicularity condition , `m_(1)m_(2)=-1`]
Also, the perpendicular bisector passing through the mid - point of the line segment joining the points A(-1,1) and B(3,3).
`:.` Mid - point `=((-1+3)/(2),(1+3)/(2))=(1,2)`
[since, mid point of the line segment joining the points`(x_(1),y_(1))and (x_(2),y_(2))`is `((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2))]`
Now , equation of perpendicular bisector have slop (-2) and passes through the point (1,2) is
`(y-2)=(-2)(x-1)`
`rArr y-2=-2x+2`
`rArr 2x+y=4` " " (i)
[since , the equation of line is `(y-y_(1))=m(x-x_(1))]`
If the perpendicular bisector cuts the y -axis , then put `x=0` in Eq . (i) , we get
`2xx0+y=4`
`rArr y=4`
Hence , the required intersection point is (0,4).