Question

In Exercises 21–24, a, b, and c are noncollinear points in\({\mathbb{R}^{\bf{2}}}\)and p is any other point in\({\mathbb{R}^{\bf{2}}}\). Let\(\Delta {\bf{abc}}\)denote the closed triangular region determined by a, b, and c, and let\(\Delta {\bf{pbc}}\)be the region determined by p, b, and c. For convenience, assume that a, b, and c are arranged so that\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\)is positive, where\(\overrightarrow {\bf{a}} \),\(\overrightarrow {\bf{b}} \)and\(\overrightarrow {\bf{c}} \)are the standard homogeneous forms for the points.

24. Take q on the line segment from b to c and consider the line through q and a, which may be written as\(p = \left( {1 - x} \right)q + xa\)for all real x. Show that, for each x,\(det\left[ {\begin{array}{*{20}{c}}{\widetilde p}&{\widetilde b}&{\widetilde c}\end{array}} \right] = x \cdot det\left[ {\begin{array}{*{20}{c}}{\widetilde a}&{\widetilde b}&{\widetilde c}\end{array}} \right]\). From this and earlier work, conclude that the parameter x is the first barycentric coordinate of p. However, by construction, the parameter x also determines the relative distance between p and q along the segment from q to a. (When x = 1, p = a.) When this fact is applied to Example 5, it shows that the colors at vertex a and the point q are smoothly interpolated as p moves along the line between a and q.

Short answer

It is proved that \(det\left[ {\begin{array}{*{20}{c}}{\widetilde p}&{\widetilde b}&{\widetilde c}\end{array}} \right] = x \cdot det\left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde b}&{\widetilde c}\end{array}} \right]\).

Step-by-step solution

Find the determinant of the points

Consider point q on the line through b and c. Then the equation becomes\({\bf{p}} = \left( {1 - x} \right){\bf{q}} + x{\bf{a}}\).

Other columns are constant, and thedeterminant is a linear function of the first column.

Write the determinant of\(det\left[ {\begin{array}{*{20}{c}}{\widetilde p}&{\widetilde b}&{\widetilde c}\end{array}} \right]\)as shown below:

\(det\left[ {\begin{array}{*{20}{c}}{\widetilde p}&{\widetilde b}&{\widetilde c}\end{array}} \right] = det\left[ {\begin{array}{*{20}{c}}{\left( {1 - x} \right)\widetilde {\bf{q}} + x\widetilde {\bf{a}}}&{\widetilde b}&{\widetilde c}\end{array}} \right]\)

Find the determinant of the points

The determinant \(det\left[ {\begin{array}{*{20}{c}}{\widetilde p}&{\widetilde b}&{\widetilde c}\end{array}} \right] = det\left[ {\begin{array}{*{20}{c}}{\left( {1 - x} \right)\widetilde {\bf{q}} + x\widetilde {\bf{a}}}&{\widetilde b}&{\widetilde c}\end{array}} \right]\) can also be written as shown below:

\(det\left[ {\begin{array}{*{20}{c}}{\widetilde p}&{\widetilde b}&{\widetilde c}\end{array}} \right] = \left( {1 - x} \right) \cdot det\left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{q}}}&{\widetilde b}&{\widetilde c}\end{array}} \right] + x \cdot det\left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde b}&{\widetilde c}\end{array}} \right]\)

From the above relation, \(\widetilde b\) is a linear combination of \(\widetilde {\bf{a}}\), and \(\widetilde {\bf{c}}\). So, \(\left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{q}}}&{\widetilde b}&{\widetilde c}\end{array}} \right]\) is a square matrix.

As \(det\left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{q}}}&{\widetilde b}&{\widetilde c}\end{array}} \right] = 0\), the relation becomes as shown below:

\(\begin{array}{l}det\left[ {\begin{array}{*{20}{c}}{\widetilde p}&{\widetilde b}&{\widetilde c}\end{array}} \right] = \left( {1 - x} \right) \cdot \left( 0 \right) + x \cdot det\left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde b}&{\widetilde c}\end{array}} \right]\\det\left[ {\begin{array}{*{20}{c}}{\widetilde p}&{\widetilde b}&{\widetilde c}\end{array}} \right] = x \cdot det\left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde b}&{\widetilde c}\end{array}} \right]\end{array}\)

Hence, it is proved that \(det\left[ {\begin{array}{*{20}{c}}{\widetilde p}&{\widetilde b}&{\widetilde c}\end{array}} \right] = x \cdot det\left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde b}&{\widetilde c}\end{array}} \right]\).