Question

Point P $\left(0,2\right)$ is the point of intersection of $y$-axis and perpendicular bisector of the line segment joining the points A $\left(-1,1\right)$ and B $\left(3,3\right)$. State whether the following statement is true or false. Justify your answer.

Answer 1

Determine whether the given statement is true or false.

The given point is P $\left(0,2\right)$ lies on the $y$-axis and is the perpendicular bisector of $AB$.

It is known that the points on the perpendicular bisector of the line segment that connects the two places are equidistant.

So, this implies that $PA=PB$.

Apply the distance formula for the two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$: $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

The point $A$ is $\left(-1,1\right)$. The value $PA$ is as follows:

$\begin{array}{rcl}PA& =& \sqrt{{\left(-1-0\right)}^2+{\left(1-2\right)}^2}\\ & \Rightarrow & PA=\sqrt{1+1}\\ & \Rightarrow & PA=\sqrt{2}\end{array}$

The point $B$ is $\left(3,3\right)$. The value $PB$ is as follows:

$\begin{array}{rcl}PB& =& \sqrt{{\left(3-0\right)}^2+{\left(3-2\right)}^2}\\ & \Rightarrow & PB=\sqrt{9+1}\\ & \Rightarrow & PB=\sqrt{10}\end{array}$

$\therefore PA\ne PB$

Therefore, the point P$\left(0,2\right)$ does not lie on the perpendicular bisector of $AB$.

Hence, the given statement is false.