8. Barycentric coordinates for a triangle

Proposition 3. Let $P,Q,R$ be three noncollinear points in    R$^ 2$. Then every point $S$ in the plane can be written uniquely in the form $S$ $=$ $c _ 1 P + c _ 2 Q + c _ 3 R$, where $c _ 1 + c _ 2 + c _ 3$ $=$ $1$.

Definition. In the Proposition, $c _ 1, c _ 2, c _ 3$ are called the barycentric coordinates of $S$ with respect to $P,Q,R$. Let's write $(c _ 1, c _ 2, c _ {3}) _ {bary}$ for the barycentric coordinates of a point.

For example, the center of mass $C$ of the triangle $PQR$ is the obtained by averaging the three vertices, so has barycentric coordinates $( {\frac{1} {3}}, {\frac{1} {3}}, {\frac{1} {3}}) _ {bary}$. In fact, ``barycentric'' means ``weight-centered''. As another example, $P$ itself has barycentric coordinates $(1,0,0) _ {bary}$. See Figure .

Figure: Triangle and its center of mass

Barycentric coordinates are often handy when you need to do something with a triangle that treats all three vertices the same way. The points inside the triangle are easily identified:

Proposition 4. A point $S$ is inside the triangle $PQR$ (or on an edge) if and only if its three barycentric coordinates with respect to $P,Q,R$ are all nonnegative.

Recall that a convex combination is a linear combination in which the coefficients are nonnegative and have sum $1$. Thus the points inside the triangle are the convex combinations of $P$, $Q$, and $R$.

Proposition 5. The barycentric coordinates of the point    x$$ with respect to $P$, $Q$, and $R$ have the values

$c _ 1 = {\frac{\displaystyle \Delta(Q,R,\mbox{\bf x})}{\displaystyle \Delta}}$,

$c _ 2 = {\frac{\displaystyle \Delta(R,P,\mbox{\bf x})}{\displaystyle \Delta}}$,

$c _ 3 = {\frac{\displaystyle \Delta(P,Q,\mbox{\bf x})}{\displaystyle \Delta}}$,

where $\Delta$ $=$ $\Delta (Q,R,P)$ $=$ $\Delta (R,P,Q)$ $=$ $\Delta (P,Q,R)$.

Thus $c _ 3$ (for example) is obtained by scaling the two-point relational form for the line through $P$ and $Q$, so is an affine function of position itself. We could write $c _ 3 ($x$)$. The scale factor of ${{\frac{\displaystyle 1}{\displaystyle \Delta}}}$ is just what is needed to have $c _ 3 (R)$ $=$ $1$, as it should be. In particular, $c _ 3$ is zero on the line through $P$ and $Q$, and is constant on each line parallel to that line.