# area analogue for $ax_2(y_2-y_1) - (1-a)y_2(x_2-x_1)$

Consider $A = ax_2(y_2-y_1) - (1-a)y_2(x_2-x_1)$.

If $a = \frac{1}{2}$ then $A = \frac{1}{2}(x_1y_2 - x_2y_1)$, which is the area of a triangle bounded by the origin, the point $(x_1,y_1)$, and the point $(x_2,y_2)$.

If $a = 0$ then $A = y_2(x_1-x_2)$ which is the area of a rectangle bounded by the x-axis with height $y_2$ and left/right sides $x=x_1$, $x=x_2$.

If $a = 1$ then $A = x_2(y_2-y_1)$ which is the area of a rectangle bounded by the y-axis with width $x_2$ and top/bottom sides $y=y_1$, $y=y_2$.

Is there a geometric analog for values of $a$ which aren't these special cases 0, 0.5, 1?

Define $\bar x=ax_2+(1-a)x_1$ and $\bar y=ay_1+(1-a)y_2$. Notice that $\bar x$ is comprised between $x_1$ and $x_2$, while $\bar y$ is comprised between $y_1$ and $y_2$. Your expression can then be rewritten as: $$A=\bar x\, y_2 -\bar y\, x_2.$$ This can be interpreted as the difference between the areas of two rectangles, both having sides parallel to the coordinate axes and one vertex at the origin, the opposite vertex being at $(\bar x,y_2)$ for the first rectangle and at $(x_2,\bar y)$ for the second rectangle.
For $a=0$ you have $\bar x= x_1$ and $\bar y=y_2$, while for $a=1$ you have $\bar x= x_2$ and $\bar y=y_1$: in those cases the difference of the above mentioned rectangles turns out to be a rectangle itself. Notice however that for $a=1$ the value of $A$ is the opposite of the rectangle area.
I don't know if a meaningful geometric interpretation can be devised for all values of $a$: we can of course find some suitable triangle or rectangle whose area is $|A|$, but this looks largely arbitrary to me.