\(P(0,2)\) is the intersection of the y–axis and the perpendicular bisector of the segment joining \(A(-1,1)\) and \(B(3,3)\).
False.
Step 1: A point lies on the perpendicular bisector of segment AB if and only if it is equidistant from A and B.
So, to check whether \(P(0,2)\) lies on the perpendicular bisector of AB, we compute \(PA\) and \(PB\) and compare them.
Step 2: Find \(PA\) using the distance formula.
\(A(-1,1)\), \(P(0,2)\)
\(PA = \sqrt{(0-(-1))^2 + (2-1)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\)
Step 3: Find \(PB\) using the distance formula.
\(B(3,3)\), \(P(0,2)\)
\(PB = \sqrt{(0-3)^2 + (2-3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}\)
Step 4: Compare the distances.
\(PA = \sqrt{2}\) and \(PB = \sqrt{10}\).
Since \(\sqrt{2} \ne \sqrt{10}\), point \(P\) is not equidistant from A and B.
Step 5: Therefore, \(P(0,2)\) does not lie on the perpendicular bisector of AB.
Also, \(P(0,2)\) is on the y-axis because its x-coordinate is 0, but it is not the intersection point of the y-axis with the perpendicular bisector.
Conclusion: The statement is False.