I am a programmer, but not much of a mathematician. How do I write the following formula?

Suppose point A has coordinates (ax, ay), and point B has coordinates (bx, by). Then all points on line CD will match

.. CD = (s*(by-ay)+ax, s*(ax-bx)+ay)

for some scalar value of "s". Positive values of "s" will define points in the direction of C, and negative values will define points in the direction of D.

We can similarly define points on line EF.

.. EF = (t*(by-ay)+bx, t*(ax-bx)+by) ... positive "t" goes in the direction of E, negative "t" in the direction of F

When s and t take on the values ±1/2, you will define four points C,D,E,F that are the corners of a square with AB joining midpoints of opposite sides. Of course there is no requirement that these have any value in particular. The larger they are, the bigger your polygon will be.

I think Postscript would be perfect for this. You have really done the math when you made the sketch.

http://paulbourke.net/dataformats/postscript/

Units are 1/72 inch. Here's the sample code to draw two connected line segments.

newpath

100 200 moveto

200 250 lineto

100 300 lineto

2 setlinewidth

stroke

Stroke means to draw the line with line width of 2. Fill would fill the rectangle.

If you know the coordinates of A (a1, A2) and D (D1, D2) , and the difference in the x coordinate of the endpoints of line AB is u and the difference in the y coordinates is v. Let A be at 0,0. and the coordinates of E and F are easy to compute. E = u,v, F = D1+u, D2+v,

Postscript would define the lines directly:

newpath

0,0 lineto

u,v lineto

D1+u, D2+v

lineto D1,D2

lineto 0,0

stroke

Rem: or fill.

I don't have a Postscript interpreter to test this, so you may have to change something. This code could print a rectangle directly on any postscript enabled device.

Good luck!

Let A be (x1,y1) B be (x2,y2).

The slope of the line between them is m=(y2-y1)/(x2-x1). The line perpendicular to AB at A has slope n=-1/m. The point slope equation of that line is (y-y1)/(x-x1)=n [1]. The equation of a circle with center at A and radius R has equation (x-x1)^2+(y-y1)^2=R^2 [2]. Find where eq 1 and eq 2 intersect. Substitute eq 1 into eq 2 and solve for x and y.

Use the dot-product relationships.

https://en.wikipedia.org/wiki/Dot_product

Define two vectors (using your picture) as follows:

X = vector connecting points A & B

Y = vector connecting points A & D

||X|| = distance between A & B

||Y|| = distance between A & F

Want point F such that X*Y = ||X||*||Y||*cos (angle) = 0. This occurs only when angle = pi / 2 relative to the line segment (A, B)