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Shrunk Axes

Date: 01/20/97 at 10:31:59
From: Jen
Subject: Coordinate axes

When using the distance formula, suppose that the coordinate axes 
are not perpendicular, but they form some other angle (say 60 deg). 
How would you find the distance between points in this system?  I was 
thinking about using Law of Sines or Cosines, but I am not really 

Date: 01/20/97 at 13:34:58
From: Doctor Ken
Subject: Re: Coordinate axes

Hi there -

To answer this question, you've got to go back and do some thinking 
about what you mean by the question.  The reason is that when we use 
coordinate axes, we essentially *define* the axes to be perpendicular.  
This question gets answered in an area of math called "Linear 
Algebra," which is usually taught in the first couple of years of 
college.  The answer is that there are a few different ways to measure 
the distance, depending on what you meant by the question, and they 
give you different results!  

But here's one way: just pretend you're in the regular system.  
For instance, if we have these points:

               2-/-                *B = (4,2)
            1-/- *A = (1,1)           
          /      1     2     3     4     5     6     7     8

See how I measured the coordinates by using a grid parallel to the 
two axes?  So once you have these coordinates, you can find the 
distance *relative to this coordinate system* by using the regular old 
distance formula. Here we have Sqrt{(4-1)^2 + (2-1)^2} = Sqrt{10} as 
the distance.

If we wanted to find the distance between these points in the 
*regular* coordinate system, we'd first have to find the coordinates 
of these points in the regular system, and then use the distance 

You can find more information about this in the College Linear Algebra 
section of our archives at:   

-Doctor Ken,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Linear Algebra

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