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

Solution:

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).