# Finding the Vertex Coordinates of a Rectangle In Cartesian Space

1. Aug 20, 2009

I am hoping to find the coordiantes of all 4 vertices when the rectangle is in any orientaion knowing the length l, the width b, the coordinate of its center mark (xcen,ycen), and the coordinate of vertex A as shown below:

This is NOT HOMEWORK so although I think it is possible to do, I am not sure that it is.

Any ideas?

I am thinking of using the vector that points from A to the center somehow... I know that if I double its length then I have arrived at the vertex C.... but how to extract those coordinates, I cannot see.

EDIT: Here is a drawing that better illustrates what I am thinking. The Blue Rectangle is the one I want to to find the vertices for. I know all information in blue.

The Black Rectangle shares the same A vertex and is in what I have DEFINED to be standard reference position (SRP).

I could find the angle of the vector that points from A to the center rc of the black rectangle and compare it with the angle of that of the blue rectangle r'c

I know that the difference $\theta - \theta '$ should be the angle that all vertices should carve out. I just can't see how to make the actual calculations of their cartesian coordinates?

Last edited: Aug 20, 2009
2. Aug 20, 2009

Looking around online and a Rotation Matrix seems promising.

Though I am finding a lot of definitions of a Rotation Matrix, I am not finding many practical examples... so I am not exactly sure what it does.

3. Aug 20, 2009

Okay Here is where I am at: Updated Diagram for reference:

rc and r'c are the vectors from A to the centers.

In order for all vertices to get from the standard ref posotion to the new positions, they must all rotate through the angle phi correct?

If A=A' is locate at the point (x0, y0), then we have the points

A(x0, y0)
B(x0, (y0+b)) *Taking down as +Y and Right as +X
C((x0+L), (y0+b))
D((x0+L), y0)

Now to use a Rotation Matrix to get the new coordinates of A' B' C' D'
I am a little confused.

Do I use the coordinates? Or the vector components of rBA,

FOR EXAMPLE: If I am looking at the vectors rBA and r'BA

And I know that to get from rBA to r'BA I rotated through the angle Phi. How do I get the new coordiantes of B out of the deal?

4. Aug 20, 2009

### Redbelly98

Staff Emeritus
Hi,

While this can be done using rotations, I don't think it's necessary.

We know that B is a distance b from A, and a distance rc from the center. In other words, it's at the intersection of two circles. You'd just need to set up the equations and solve them.

One problem I see, which is that there are two solutions for the location of B. You'll need another condition or some way to specify which solution is the desired one.