Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.
At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.
We parameterize the line segment by
[tex]\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t[/tex]
[tex]\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle[/tex]
with [tex]0\le t\le1[/tex]. Then we find the differential:
[tex]\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle[/tex]
[tex]\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt[/tex]
with [tex]0\le t\le1[/tex].
Here, the line integral is
[tex]\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle[/tex]
[tex]=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt[/tex]
[tex]=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt[/tex]
[tex]=x_1y_2-x_2y_1[/tex]
as required.
