My Math Forum Configuration centered at the origin

 Algebra Pre-Algebra and Basic Algebra Math Forum

 July 10th, 2010, 04:05 AM #1 Newbie   Joined: Jul 2010 Posts: 26 Thanks: 0 Configuration centered at the origin I have two configurations (set of points, each with its x and y coordinate) that should be compared. In order to do this, I should perform the operation so that both configurations are centered at the origin. I understand that the operations are to be performed on x and y axis, but I'm not sure how to do this. Any suggestions are welcome.
 July 10th, 2010, 04:28 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Configuration centered at the origin How are they to be centred? Probably you want the centroid, but it depends what the points represent.
 July 10th, 2010, 06:43 AM #3 Newbie   Joined: Jul 2010 Posts: 26 Thanks: 0 Re: Configuration centered at the origin The configurations should have centroids at the origin. I've found this, but I'm not sure what it means: "Subtract the mean vectors for the configurations from each of the respective points in order to have the centroids at the origin"
 July 10th, 2010, 07:13 AM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Configuration centered at the origin Suppose that one configuration of points is $\{(x_1,y_1), (x_2,y_2), (x_3,y_3), ... (x_n,y_n)\}$ Then the centroid will be at $(\bar{x},\bar{y})= \left(\frac{x_1+x_2+...+x_n}{n},\quad\frac{y_1+y_2 +...+y_n}{n}\right)$ So to place the centroid at the origin, subtract that centroid's coordinates from each point's coordinates: $\{(x_1-\bar{x},y_1-\bar{y}), (x_2-\bar{x},y_2-\bar{y}), (x_3-\bar{x},y_3-\bar{y}), ... (x_n-\bar{x},y_n-\bar{y})\}$ For example take the triangle with vertices {(2,3), (5,-1), (3,). The centroid is at ((2+5+3)/3, (3+(-1)+/3) = (10/3, 10/3), so subtract 10/3 from every coordinate to centre the triangle at the origin.

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