Parametric Form for Equation of a LineDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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