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.

23. Let p be any point in the interior of\(\Delta {\bf{abc}}\), with barycentric coordinates\(\left( {r,s,t} \right)\), so that

\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}r\\s\\t\end{array}} \right] = \widetilde {\bf{p}}\)

Use Exercise 21 and a fact about determinants (Chapter 3) to show that

\(r = \left( {area of \Delta pbc} \right)/\left( {area of \Delta abc} \right)\)

\(s = \left( {area of \Delta apc} \right)/\left( {area of \Delta abc} \right)\)

\(t = \left( {area of \Delta abp} \right)/\left( {area of \Delta abc} \right)\)

Short answer

It is proved that \(r = \left( {area of \Delta pbc} \right)/\left( {area of \Delta abc} \right)\), \(s = \left( {area of \Delta apc} \right)/\left( {area of \Delta abc} \right)\), and \(t = \left( {area of \Delta abp} \right)/\left( {area of \Delta abc} \right)\).

Step-by-step solution

State the equation of point p

Write the equation for point p as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}r\\s\\t\end{array}} \right] = \widetilde {\bf{p}}\\r\widetilde {\bf{a}} + s\widetilde {\bf{b}} + t\widetilde {\bf{c}} = \widetilde {\bf{p}}\end{array}\)

Use the Crammer rule

Use theCrammer rule to solve for r.

\(r = \frac{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{p}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}\)

By the concept given in Exercise 21, \(\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{p}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]\)is twice the area of\(\Delta {\bf{pbc}}\).

It can be written as shown below:

\(\begin{array}{c}r = \frac{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{p}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}\\ = \frac{{2\left[ {{\rm{ar}}\left( {\Delta {\bf{pbc}}} \right)} \right]}}{{2\left[ {{\rm{ar}}\left( {\Delta {\bf{abc}}} \right)} \right]}}\\ = \frac{{{\rm{area of }}\Delta {\bf{pbc}}}}{{{\rm{area of }}\Delta {\bf{abc}}}}\end{array}\)

Thus, it is proved that\[r = \left( {area of \Delta pbc} \right)/\left( {area of \Delta abc} \right)\].

Use the Crammer rule to solve for s.

\(s = \frac{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{p}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}\)

By the concept given in Exercise 21, \(\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{p}}}&{\widetilde {\bf{c}}}\end{array}} \right]\)is twice the area of\(\Delta {\bf{apc}}\).

It can be written as shown below:

\(\begin{array}{c}s = \frac{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{p}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}\\ = \frac{{2\left[ {{\rm{ar}}\left( {\Delta {\bf{apc}}} \right)} \right]}}{{2\left[ {{\rm{ar}}\left( {\Delta {\bf{abc}}} \right)} \right]}}\\ = \frac{{{\rm{area of }}\Delta {\bf{apc}}}}{{{\rm{area of }}\Delta {\bf{abc}}}}\end{array}\)

Thus, it is proved that\[s = \left( {area of \Delta apc} \right)/\left( {area of \Delta abc} \right)\].

Use the Crammer rule to solve for t.

\(t = \frac{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{p}}}\end{array}} \right]}}{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}\)

By the concept given in Exercise 21, \(\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{p}}}\end{array}} \right]\)is twice the area of\(\Delta {\bf{abp}}\).

It can be written as shown below:

\(\begin{array}{c}s = \frac{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{p}}}\end{array}} \right]}}{{\det \left[ {\begin{array}{*{20}{c}}{\widetilde {\bf{a}}}&{\widetilde {\bf{b}}}&{\widetilde {\bf{c}}}\end{array}} \right]}}\\ = \frac{{2\left[ {{\rm{ar}}\left( {\Delta {\bf{abp}}} \right)} \right]}}{{2\left[ {{\rm{ar}}\left( {\Delta {\bf{abc}}} \right)} \right]}}\\ = \frac{{{\rm{area of }}\Delta {\bf{abp}}}}{{{\rm{area of }}\Delta {\bf{abc}}}}\end{array}\)

Thus, it is proved that \[t = \left( {area of \Delta abp} \right)/\left( {area of \Delta abc} \right)\].