# Parameterising a line given by two points

1. Jun 19, 2013

### mrcleanhands

I'm working on a green's theorem integral and I'm given a line (0,0) to (1,2) which i parameterise as (t,2t) but the answer is actually (1-t, 2(1-t)) and if I use my parameterisation on the integral I get a different answer!

2. Jun 19, 2013

### Office_Shredder

Staff Emeritus
The exact parametrization of the integral isn't relevant, but you have to integrate in the correct direction. Your parametrization starts at (0,0) when t=0 and moves to (1,2) when t=1, the other parametrization starts at (1,2) when t= 0 and moves to (0,0) when t=1. Green's theorem includes a statement about which direction along the boundary you are supposed to integrate

3. Jun 20, 2013

### HallsofIvy

Staff Emeritus
Using (t, 2t), when t= 0 you get (0, 0) and when t= 1 you get (1, 2). Using (1- t, 2(1- t)), when t= 0 you get (1, 2) and when t= 1 you get (0, 0). Those both give exactly the same line, just oriented oppositely.

The distinction is the orientation of the line. You cannot just integrate 'on' the line, you must either integrate from (0, 0) to (1, 2) or from (1, 2) to (0, 0). If, using your parameterization, you integrate $\int_0^1 f(x(t), y(x))dt$ you are integrating from (0, 0) to (1, 2). If, using the other parameterization, you integrate $\int_0^1 f(x(t),y(t))dt$, you are integrating from (1, 2) to (0, 0) and will get the negative[/b[] of the previous result. You must think about which direction you want to go on the line and choose your limits of integration accordingly. If you want to integrate from (0, 0) to (1, 2) using the x= 1- t, y= 2(1- t) parameterization, since t= 1 gives (0, 0) and t= 0 gives (1, 2), you must integrate from t= 1 to t= 0: $\int_1^0 f(x(t), y(t))dt$.