3. The two-point relational form

Suppose we want to find a relational description $f(x,y) = 0$ for the line through two given points $P$ and $Q$. There is a particular choice of $f$ that does this nicely. It can be derived in either of two ways.

For the first way, take any third point $R$, and recall the affine transformation that takes the standard triangle to $P,Q,R$. Let the determinant of its extended matrix be $\Delta (P,Q,R)$. By adding the third row to the first two, we get a simpler description. Thus

$\Delta (P,Q,R)$ $=$ $\det {\left[\begin{array}{cc}(P-R)&0\\ (Q-R)&0\\ R&1\end{array}\right]}$ $=$ $\det {\left[\begin{array}{cc}P&1\\ Q&1\\ R&1\end{array}\right]}$ $=$ $\det { \left[\begin{array}{ccc}p _ 1& p _ 2& 1\\ q _ 1& q _ 2& 1\\ r _ 1& r _ 2& 1\end{array}\right]}$.

Here are some facts about $\Delta (P,Q,R)$. You have seen the first three; the last two follow from (a).

Proposition 1. For any points $P,Q,R$ in    R$^ 2$:

(a) the area of the triangle formed by $P,Q,R$ is ${\frac{1} {2}} \Delta (P,Q,R)$ in absolute value;

(b) if $\Delta (P,Q,R) > 0$ then $P,Q,R$ have the same orientation as the standard triangle, namely, as you go from $P$ to $Q$ to $R$ and back to $P$ you are traversing the triangle counterclockwise; equivalently, $Q$ is to the left of $P$ as seen from $R$, just as $(0,1)$ is to the left of $(1,0)$ as seen from the origin;

(c) similarly, if $\Delta (P,Q,R) < 0$, then the traversal $P$ to $Q$ to $R$ to $P$ is clockwise; equivalently, $Q$ is to the right of $P$ as seen from $R$;

(d) $\Delta (P,Q,R) = 0$ when $P,Q,R$ are collinear (lie on a line) and so form only a degenerate triangle of zero area;

(e) $\Delta (P,Q,R)$ $=$ $\Delta (Q,R,P)$ $=$ $\Delta (R,P,Q)$ (i.e., $\Delta$ doesn't change when the three points are permuted cyclically).

By (d), the points $R$ for which $\Delta (P,Q,R)$ $=$ $0$ form the line we're looking for. To emphasize which point is varying, let's put    x$$ for $R$:

Proposition 2. The line $L$ in    R$^ 2$ through two points $P$ and $Q$ has equation

$\Delta (P,Q,$x$)$ $=$ $0$.

Let's call this the two-point relational form of $L$. To see that the function $\Delta (P,Q,$x$)$ is really affine, expand $\Delta (P,Q,$x$)$ by cofactors of the third row; you get

$\Delta (P,Q,$x$)$ $=$ $\det { \left[\begin{array}{ccc}p _ 1& p _ 2& 1\\ q _ 1& q _ 2& 1\\ x _ 1& y _ 2& 1\end{array}\right]}$ $=$ $(p _ 2 - q _ {2}x) + (q _ 1 - p _ {1}y) + (p _ 1 q _ 2 - p _ 2 q _ {1})$.

The second way of deriving the two-point relational form is to use $P$ and $Q$ to make a normal to the desired line $L$:

For the the normal use $Q-P$ rotated counterclockwise $90 ^\circ$. Thus if $P = (p _ 1, p _ {2})$ and $Q = (q _ 1, q _ {2})$, let $[a,b]$ $=$ $(q _ 1 - p _ 1, q _ 2 - p _ {2}) {\left[\begin{array}{cc}0&1\\ -1&0\end{array}\right]}$ $=$ $[-(q _ 2 - p _ {2}),q _ 1 - p _ {1}]$. Since $P$ is on $L$, $a p _ 1 + b p _ 2 + c$ $=$ $0$, so $c = - (a p _ 1 + b p _ {2})$ $=$ $p _ 1 q _ 2 -p _ 2 q _ 1$. Thus our affine function is

$(p _ 2 - q _ {2}x) + (q _ 1 - p _ {1}y) + (p _ 1 q _ 2 - p _ 2 q _ {1})$, as before.

Note 1. From the second method of deriving the two-point relational form, it is clear that if you walk along the line $L$ from $P$ towards $Q$, the half-plane with $\Delta (P,Q,$x$) > 0$ will be on your left. This is also clear from (b) of Proposition 1, if you put $R$ $=$    x$$ again.

Note 2. Because the two-point relational form does not involve division, it is especially good for use in a computer without built-in floating-point operations, if you use only integers for coordinate values of points.