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I want to create a rectangular polygon using two points as guides.

So let's say a journey starts in Egypt and ends in London, my polygon should have 4 points:

  • 10 miles further from London than Egypt is, following the line between them (roughly south in this example).
  • Halfway between the two cities but 50 miles at right angles from the line that joins them.
  • Like the above point but 50 miles in the other direction.
  • 20 miles further from Egypt than London is, following the line between them (roughly north in this example).

I'll end up with a rough diamond shaped polygon that would completely contain the straight line journey from Egypt to London.

I hope this makes sense; any help for how I can calculate the 4 points is appreciated.

enter image description here

I have also asked this on gis.stackexchange.com but as I only need 2D and the fact that its map coordinates is irrelevant. I thought I would ask it here as well.

EDIT:

Thanks to Ju'x i have come a long way, but the outer points dont seem to be at 90 degrees to the mid point (image here: https://i.sstatic.net/lxX9l.png). I am pretty sure i have followed the answer given correctly....

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Let's say $L$ with coordinates $(x_L;y_L)$ stands for London, and $C$ with coordinates $(x_C; y_C)$ stands for Cairo (Egypt is hardly a point on a map). I will assume that the coordinates are expressed in miles, so that (for example) $$ LC = \sqrt{(x_C-x_L)^2 + (y_L - y_c)^2} $$ is the distance between London and Cairo in miles.

The unit vector $u = \dfrac{\overrightarrow{LC}}{\left\|\overrightarrow{LC}\right\|}$ has coordinates $\left(\dfrac{x_C-x_L}{LC},\dfrac{y_C-y_L}{LC}\right) = \left(x_u,y_u\right)$.

The two extremal points of your polygon are $C' = C + 10u$ and $L' = L - 20u$ : $$ C' = \left(x_C + 10x_u,y_c + 10y_u\right)\qquad L' = \left(x_L - 20x_u, y_L - 20y_u\right) $$

The midpoint $M$ has coordinates $(x_M,y_M)=\left(\dfrac{x_C+x_L}{2},\dfrac{y_C+y_L}{2}\right)$, and an orthogonal vector to $u$ is the unit vector $v$ with coordinates $(x_v,y_v) = (y_u, -x_u)$. The two lateral points of your polygon are $A = M + 50v$ and $B = M - 50v$ : $$ A = \left(x_M + 50x_v, y_M + 50 y_v\right),\qquad B = \left(x_M - 50x_v, y_M - 50y_v\right). $$

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  • $\begingroup$ Thanks so much, i dont seem to be creating the 2 lateral points correctly though, i have added a new image link above and here i.sstatic.net/lxX9l.png to demonstrate. $\endgroup$ Commented Jan 16, 2013 at 13:51
  • $\begingroup$ You should check $v$: if $u = (a,b)$ then take $v=(b, -a)$. $\endgroup$ Commented Jan 16, 2013 at 13:56
  • $\begingroup$ you sir, are both a scholar and a gentleman.. thank you. $\endgroup$ Commented Jan 16, 2013 at 15:42
  • $\begingroup$ @Siméon maybe you can help me with this question: math.stackexchange.com/questions/1548841/… $\endgroup$ Commented Nov 27, 2015 at 16:38

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