Shorter way :


A square has 4 vertices.

Every 2 adjacent sides share a common vertex, are mutually perpendicular and have equal lengths/measures.


Join (0,8) to (1,3) and join (6,4) to (1,3) with lines.


The lines meet at a common point (1,3)

Since it's already given that (0,8) and (6,4) are diagonally opposite vertices of a square,

if we can show that the length of both lines are equal, and the two lines are perpendicular to each other, then we can say that they are adjacent sides of a square. Then this implies that (1,3) is one of its vertices.

Using the formula for distance between two points,


Distance/length between (0,8) and (1,3)
= √((0 - 1)² + (8 - 3)²) units
= √(1 + 25) units
= √26 units

Distance/length between (6,4) and (1,3)
= √((6 - 1)² + (4 - 3)²) units
= √(25 + 1) units
= √26 units

So the two lengths are equal.

Next,

Gradient of line joining (0,8) and (1,3)

= (3 - 8) / (1 - 0)
= -5/1
= -5

Gradient of line joining (6,4) and (1,3)

= (3 - 4) / (1 - 6)
= -1/-5
= 1/5

Product of their gradients = (-5)(1/5) = -1

Since this is -1, the two lines are mutually perpendicular.

ii)

Midpoint of diagonal = (3,6)

(see earlier working in the main comments section)

Let (a,b) be the coordinates of the fourth vertex


Now the line joining (a,b) and (1,3) is the other diagonal. The midpoint (3,6) is common to both diagonals.


Midpoint also
= ((a + 1)/2, (b + 3)/2)
= (3,6)


So,


(a + 1)/2 = 3
a + 1 = 6
a = 5


(b + 3)/2 = 6
b + 3 = 12
b = 9

The other vertex is (5,9)

iii)

Area of square = (length of square)²

= (√26 units)²

= 26 units²