Co-ordinate Geometry — Maths STD 10 — Question

The distance d between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula,
$\text{d}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2}$
In a square all the sides are equal to each other.
Here the four points are $A(5, p), B(1, 5), C(2, 1)$ and $D(6, 2)$.
The vertex $'A'$ should be equidistant from $'B'$ as well as $'D'$.
Let us now find out the distance $'AB'$ and $'AD'$.
$\text{AB}=\sqrt{(5-1)^2+(\text{p}-5)^2}$
$\text{AB}=\sqrt{(4)^2+(\text{p}-5)^2}$
$\text{AD}=\sqrt{(5-6)^2+(\text{p}-2)^2}$
$\text{AD}=\sqrt{(-1)^2+(\text{p}-2)^2}$
These two need to be equal.
Equating the above two equations we have,
$AB = AD$
$\sqrt{(4)^2+(\text{p}-5)^2}=\sqrt{(-1)^2+(\text{p}-2)^2}$
Squaring on both sides we have,
$ (4)^2+(p-5)^2=(-1)^2+(p-2)^2 $
$ 16+p^2+25-10 p=1+p^2+4-4 p $
$ 6 p=36 $
$ p=6$