# Closed Line Integral

1. May 13, 2012

### nayfie

1. The problem statement, all variables and given/known data

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2. Relevant equations

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3. The attempt at a solution

This isn't really a proper homework question so I'll just write my problem here:

I'm trying to compute a closed line integral over a triangular region. I have calculated two of the sides, but am now left with the hypotenuse. I'm trying to find a parametrisation for this to work.

My lecturer told me that the equation for a line is:

$((1 - t)(a, b) + t(c, d))$

Yet I had always known the equation of a line to be:

$(a, b) + t(c, d)$

I have tried computing this line integral using both parametrisations. The first one works, but the second one doesn't. (They both give the same value but the first one is negative and the second one is positive. For the line integral to give what I believe is the correct answer, the line integral for this section needs to be negative).

Can somebody explain why the first equation works but the second one doesn't? Also can you please give some insight into how the first one works? I don't understand where the $(1 - t)$ at the beginning comes from.

2. May 13, 2012

### tiny-tim

hi nayfie!
the first is for the line from (a,b) to (c,d)

the second is for the line from (a,b) and parallel to the line from (0,0) to (c,d)

3. May 13, 2012

### HallsofIvy

Staff Emeritus
Specifically, with (1- t)(a, b)+ t(c, d) when t=0, that is (a, b) and when t= 1 that is (c, d) so that is the line through points (a, b) and (c, d). With (a, b)+t(c, d) when t= 0, that is (a, b) and when t= 1 it is (a+c, b+ d). The "slope" is (b+d- (b))/(a+b- (a))= d/b, the same as the first line so they are parallel as tiny-tim said, but, in general, does not contain the point (c, d). (It will contain (c, d) if and only if we can solve a+ tc= c, b+ dt= d for the same value of t. That is, we must have (c-a)/c= (d- b)/d.)

The form you give is, as you say, a line but, in general, not the specific line you want.

4. May 13, 2012

### nayfie

Thank you both for the replies, very helpful!