The Long way :


The square can be divided into four congruent isosceles right-angled triangles by drawing two diagonals connecting each pair of diagonally opposite vertices.

The two diagonals intersect at their common midpoint, which is the centre of the square.


Each of the four triangles have 2 perpendicular sides that are equal in length


This length = half a diagonal's length

= ½√((6-0)²+(4 - 8)²) units
(using the formula for distance/length between two points)

= ½√(36 + 16) units
= ½√52 units
= √13 units


Next,

Coordinates of the square's centre
= Coordinates of the midpoint of the diagonal joining (6,4) to (0,8)

= ((6 + 0)/2 , (4 + 8)/2)
= (3,6)

Gradient of the line this diagonal lies on
= (8 - 4)/(0 - 6)
= 4/-6
= -⅔

Since the two diagonals are perpendicular to each other, the product of their gradients = -1


Gradient of the line the other diagonal lies on
= -1/-⅔
= 3/2

Sub (3,6) and m = 3/2 into the general form of the equation of a line y = mx + c

6 = 3/2 (3) + c
6 = 9/2 + c
c = 3/2

So the other diagonal lines on the line y = 3/2 x + 3/2

The other two vertices lie on this line.

Let (x,y) be the possible coordinates of one of those vertices.


Length of each triangle's perpendicular sides

= distance between midpoint (3,6) and (x,y)
= √((x - 3)² + (y - 6)²)

Then √((x - 3)² + (y - 6)²) = √13
(see earlier working)


Square both sides,

(x - 3)² + (y - 6)² = 13

(Side note : this is actually the equation of a circle that the square is inscribed in.)


As the vertices lie on the line, their coordinated satisfy the equation y = 3/2 x + 3/2

Sub y = 3/2 x + 3/2 ,

(x - 3)² + (3/2 x + 3/2 - 6)² = 13

x² - 6x + 9 + (3/2 x - 9/2)² = 13

x² - 6x + 9 + (9/4 x² - 27/2 x + 81/4) = 13


Multiply both sides by 4 ,

4x² - 24x + 36 + (9x² - 54x + 81) = 52

13x² - 78x + 117 = 52

13x² - 78x + 65 = 0

Divide both sides by 13,

x² - 6x + 5 = 0

Factorise ,

(x - 1)(x - 5) = 0

x - 1 = 0 or x - 5 = 0

x = 1 or x = 5


Then,

y = 3/2 (1) + 3/2 or y = 3/2 (5) + 3/2

y = 3/2 + 3/2 or y = 15/2 + 3/2

y = 3 or y = 9


So the coordinates of the other two vertices are (1,3) and (5,9)

(Shown)

Length of 1 side of a square

= Length joining (0,8) to (1,3)

(or you can use (0,8) to (5,9) , (6,4) to (5,9) , (5,9 to (1,3) )

= √((1 - 0)² + (3 - 8)²)

= √(1 + 25)

= √26 units


Area of square

= (Length of one side)²

= (√26 unit)²

= 26 unit²


Alternatively,


Use Pythagoras' Theorem on a diagonal and two sides of the square


(Diagonal length)² = (length of square)² + (length of square)²


52 = 2 (length of square)²

26 = (length of square)² = area of square

Area of square = 26 square units

Shorter way (probably what the question expects)


If (1,3) is the coordinates of one of the remaining vertices,

So technically, for the shorter way, i could just assume that one of the vertices is (1,3) and use it to prove? Like using pythagorus theorem and the length of the line and prove they are perpendicular using their gradients?

Because I’m not sure if I can straight away assume that the point is (1,3).

The shorter way is not assuming (1,3) is a vertex.


It's making observations from existing information, then using these observations to come up with inferences and deductions, to arrive at the conclusion that (1,3) is a vertex.


Notice that in the working we did not actually say (1,3) is a vertex.


See the working on the answer section.

I noticed you mentioned Pythagoras' Theorem, which I didn't apply in my working at the other side.


This approach would be :


length of line joining (0,8) and (1,3)
= √26 units

length of line joining (6,4) and (1,3)
= √26 units

The two lengths are equal.


Next, make the observation that :

(length of line joining (6,4) and (1,3)) ² + (length of line joining (0,8) and (1,3))²


= (√26 units)² + (√26 units)²
= 26 units² + 26 units²
= 52 units²
= (√52 units)²
= (length of diagonal joining (0,8) and 6,4)²


This satisfies the Pythagoras' Theorem a² + b² = c² .

Since the 3 lines are sides of the same triangle, (the 3 lines are connected at the 3 points, which forms a triangle)

We deduce that :

① This triangle is a right-angled triangle.
② The diagonal of the square is the triangle's hypotenuse.
③ The other two sides have equal lengths, so this triangle is an isosceles triangle.

④ Since the hypotenuse is directly opposite the right angle, then the right angle is the angle formed between the other two sides.

This means that the two sides are perpendicular to each other.


And ,

Since (0,8) and (6,4) are diagonally opposite vertices of a square,

Then this right-angled triangle must be one half of the square.

Which leads to the deduction that (1,3) is a vertex of the square.

Got it. Thanks. ☺️

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