Point P(0, 2) is the point of intersection

False

We know that, the points lies on perpendicular bisector of the line segment joining the two

points is equidistant from these two points.

$\therefore$ $P A=\sqrt{[-4-(4)]^{2}+(6-2)^{2}}$

$=\sqrt{(0)^{2}+(4)^{2}}=4$

$P B=\sqrt{[-4-4]^{2}+(-6-2)^{2}} \sqrt{(0)^{2}+(-8)^{2}}=8$

$\because$ $P A \neq P B$

So, the point $P$ does not lie on the perpendicular bisctor of $A B$. Alternate Method

Slope of the line segment joining the points $A(-1,1)$ and $B(3,3), m_{1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}$

$\left[\because m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right]$

Since, the perpendicular bisector is perpendicular to the line segment, so its slope,

$m_{2}=-\frac{1}{\left(y_{2}\right)}=-2$

[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 $B(3,3)$.

$\therefore$ Mid-point $=\left(\frac{-1+3}{2}, \frac{1+3}{2}\right)=(1,2)$

[since, mid-point of the line segment joining the points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is 

$\left.\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\right]$

Now, equation of perpendicular bisector have slope $(-2)$ and passes through the point $(1,2)$ is

$(y-2)=(-2)(x-1)$

$\Rightarrow$ $y-2=-2 x+2$

$\Rightarrow \quad 2 x+y=4$ $\ldots$ (i)

[since, the equation of line is $\left.\left(y-y_{1}\right)=m\left(x-x_{1}\right)\right]$

If the perpendicular bisector cuts the $Y$-axis, then put $x=0$ in Eq. (i), we get

$2 \times 0+y=4$

$\Rightarrow \quad y=4$

Hence, the required intersection point is (0, 4).