Parametric Form for Equation of a Line

Date: 6/30/96 at 23:45:9
From: Anonymous
Subject: Parametric Form for Equation of a Line

Dr. Math, how can you convert an equation such as:
		
	y = -3x/4 + 7/2

to parametric vector form?

In fact, how is it done generally?

Date: 7/1/96 at 7:37:40
From: Doctor Anthony
Subject: Re: Parametric Form for Equation of a Line

The way to get the equation into parametric form is to write it as 
follows:

   (x-x1)/a = (y-y1)/b = t   where (x1,y1) is a point on the line, a 
and b are called direction cosines, and t is the parameter.  Then we 
have:

 x = x1 + a*t     y = y1 + b*t 

To get the equation you quoted into this form, we first get rid of 
denominators by multiplying through by 4.

    4y = -3x +14.  Now find a convenient point on the line to 
represent (x1,y1).  You can give x any value and then find 
corresponding y value, but it is preferable to choose an x value which 
leads to a nice looking equation.  If I let x = 2, then 
4y = -6+14 = 8, giving y=2 and so (2,2) is a point on the line.  The 
equation of the line can now be written:

    4(y-2) = -3(x-2)  i.e. we replace x by (x-2) and y by (y-2) and 
forget the constant term which has now been absorbed into the other 
terms of the equation.

      So we have -3(x-2) = 4(y-2)

                 (x-2)/4 = (y-2)/-3 = t

We can put (x-2)/4 = t,  x-2 = 4t,    x = 2 + 4t
   also   (y-2)/-3 = t,  y-2 = -3t,   y = 2 - 3t
     
And so the parametric form of the line is  x = 2 + 4t
                                           y = 2 - 3t 

-Doctor Anthony,  The Math Forum
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