Jacobian transformation for finding area

[Titan97]

My maths teacher taught me a shortcut for finding area bounded by curves of the form: $$|as+by+c|+|Ax+By+C|=d$$

Shortcut:

Let required area be $A_0$ and new area after "transformation" be $A$
Then, $$A_0\begin{vmatrix}
a& b\\
A& B\end{vmatrix}=A=2d^2$$

All I understood was the $A=2d^2$ part. Its the area of triangle of base=y-intercept and height=x-intercept where x_intercept is c/a and y-intercept is b/a.

I have not even heard the name "jacobian" and I don't know what transformation he was talking about. But the formula worked. I want to learn about Jacobian (the transformation and not the person). How did he get the shortcut? (I did not understand what's given in wikipedia and they have not specified this shortcut)

 

[andrewkirk]

Let $f(x,y)\equiv ax+by+c$ and $F(x,y)\equiv Ax+By+C$. Consider the lines on the number plane that are the solutions of the equations $f(x,y)=nd$ for $n$ an integer, and the lines that are the solutions of $F(x,y)=md$ for $m$ an integer. The $F$ lines will all be parallel to one another and equidistant, and the same for the $f$ lines. The lines give a lattice for the number plane that can be used for an alternative coordinate system under which a point at the intersection of lines $f(x,y)=nd$ and $F(x,y)=md$ is given coordinates $(n,m)$, and coordinates for points not on such lines are interpolated.

The area to be measured is that of the parallelogram whose vertices have alternative (n,m) coordinates (0,1), (1,0), (0,-1), (-1,0). In the original coordinates the area of such a shape (which would be a square) is 2 but because the f and F lines are probably not orthogonal, and are scaled differently, the shape will be a parallelogram. The Jacobian is a matrix that, amongst other things, represents the impact of a change in coordinates. It can be used to relate the area of a parallelogram to the area of a square in the coordinate system in which the parallelogram has 'square' coordinates.

If you search 'jacobian matrix measure area parallelogram' you'll find lots of good info about it.

In two dimensions this can be done more easily with vector cross products, but the Jacobian approach allows application to any number of dimensions.