$(x_1y_2 - x_2y_1)$: area or displacement?

Question

Whenever I run into an expressions of the form $x_1y_2 - x_2y_1$, I usually have no problem finding some pertinent way to interpret it as the area of the parallelogram subtended by some two vectors.

But here's a case in which I can't figure out how to apply such an interpretation.

The line passing through points $[x_1, y_1]$ and $[x_2, y_2]$ consists of all those points $[x, y]$ that satisfy the equation

$$(y_1 - y_2)\,x + (x_2 - x_1)\,y + (x_1y_2 - x_2y_1) = 0$$

Now, in expressions of the form $A\,x + B\,y + C = 0$, I always interpret the $C$ term as a "location parameter", i.e. something "one-dimensional" somehow. And yet, if $C = (x_1y_2 - x_2y_1)$, as the above expression says, it seems to corresponds more to an area.

How can these two seemingly contradictory intuitions be reconciled?

Answer 1

You're seeing a glimpse of projective duality between points and lines in the projective plane $\mathbb{P}^2$. A line through the points $(x_1;y_1;z_1)$ and $(x_2;y_2;z_2)$ in $\mathbb{P}^2$ has equation $$(y_1 z_2 - y_2 z_1) x + (x_2 z_1 - x_1 z_2) y + (x_1 y_2 - x_2 y_1) z = 0.$$ Conversely, the point of intersection of the lines $A_1 x + B_1 y + C_1 z = 0$ and $A_2 x + B_2 y + C_2 z = 0$ is $$(B_1 C_2 - B_2 C_1, A_2 C_1 - A_1 C_2, A_1 B_2 - A_2 B_1).$$

I don't have a really good explanation of how units should work in duality. I think of it algebraically as dual spaces should have opposite units. But perhaps someone with better physical intuition than me can add something more useful.

Answer 2

Rewrite as $(xy_1 - yx_1) + (x_1y_2 - y_1x_2) = (x_2y_1 - x_1y_2)$.

This is a sum of three cross products. Each can be thought of as an area of a parallelogram or as twice the area of a triangle. So this equation says that the triangle formed by the origin, $(x_1,y_1)$, and $(x_2,y_2)$ is the union (in some directed sense) of the triangles formed by the origin and the pairs of points $((x,y),(x_1,y_1))$ and $((x,y),(x_2,y_2))$. By drawing a quick picture it should be reasonable to see that this corresponds to the three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x,y)$ being collinear.

Answer 3

If I understand rightly, the C term is a scalar. It has no direction, only magnitude. The x and y terms are displacements in the respective directions. It is possible to think of the x and y as an area (multiply the x and y direction to get the area covered). In this case, the displacements are merely components of the area. If you think of the displacements as area components (length and width), there will be no contradictions between the two concepts. Both concepts will mutually reinforce each other.

Answer 4

The quantity $x_1y_2-x_2y_1$ is indeed the (oriented) area of the parallelogram spanned by $<x_1,x_2>$ and $<y_1,y_2>$, but this doesn't mean every time this quantity pops up in some situation the associated parallelogram is going to be geometrically relevant. It's like saying each time I see the equation $y=mx^2$ I want think of the equivalence of mass and energy, but of course this type of equation pops up in all different sorts of circumstances.