Each side of a square is of length 6 units and the centre of the square is (–1 2). One of its diagonals is parallel to x + y = 0. Find the coordinates of the vertices of the square.

Let a square be ABCD

Length of sides of square = 6

Length of diagonal = 6√2

[ For a square of side a units, diagonal equals to a√2 units]

Coordinates of centre = (–1, 2)

Distance of coordinates form centre of square

Also, diagonal AC is parallel to the line, x + y = 0

So, Equation of diagonal becomes x + y = k

As the diagonal passes through the centre of the square, coordinates of centre will satisfy the equation of diagonal.

Putting points of centre in above equation, we get

(– 1) + (2) = k

K = 1

So, equation of AC becomes, x + y = 1

⇒ x = 1 – y

Let coordinate of a point of square be (t,s)

t = 1 – s

Using distance formula between the centre and a coordinate of square,

√((1–s + 1)² + (s–2)² ) = 3√2

(2–s)² + (s–2)² = 18

2(s)² + 8 –8s = 18

(s)² –4s–10 = 0

S= –1 and 5

Since slope of AC = –1

⇒ slope of BD = 1

∴ AB and CD are parallel to y–axis

And AD and BC are parallel to y axis

So, coordinates are A(2, –1), B(–4,–1),C(–4,5), D(2,5)