Finding the vertices of a square - straight lines

Question

Question: Each side of a square is of length $6$ units and the center of the square is $(-1, 2)$. One of its diagonals is parallel to $x + y = 0$. Find the co-ordinates of the vertices of the square.

What I have done: I found the equations of the two diagonals of the square, using the given line, the center point, and the fact that the diagonals are perpendicular to each other. Using Pythagoras's Theorem, I found that the distance of the center from the vertices is $3\sqrt2$.

I can obviously use the distance formula in collaboration with the equation of the line to find the four vertices of the square. However, that would be a lengthy procedure. I am sure there is an easier way to do this.

Answer 1

I think a diagram speaks for itself...

Answer 2

Will it be really lengthy?

Any point on the straight line $x+y=1$ can be written as $(h,1-h)$

So, the distance of $(h,1-h);(-1,2)$ will be the half of a diagonal

$$(h+1)^2+(1-h-2)^2=\left(\frac{6\sqrt2}2\right)^2\iff2(h+1)^2=18\iff h+1=\pm3$$