Chapter 10.6, Problem 63E

a.

To determine

The coordinates of the vertices of the surrounding rectangle and calculate the area.

Answer to Problem 63E

The area of the rectangle is $R=\left(a_3-a_1\right).\left(b_2-b_1\right)=a_3{b}_2+a_1{b}_1-a_1{b}_1-a_3{b}_2$

Explanation of Solution

First finding the coordinates by seeing how the rectangle is formed by the lines $x=a_1,x=a_3,y=b_1,y=b_2.$

So, the line equation by noting he coordinates of the points on them. So, the four vertices of the rectangle are given by,

  $\left(a_1,b_1\right),\left(a_1,b_2\right),\left(a_3,b_1\right),\left(a_3,b_2\right)$

The area of the rectangle id then simply,

  $R=\left(a_3-a_1\right).\left(b_2-b_1\right)=a_3{b}_2+a_1{b}_1-a_1{b}_1-a_3{b}_2$

Conclusion:

Hence, the area of the rectangle is $R=\left(a_3-a_1\right).\left(b_2-b_1\right)=a_3{b}_2+a_1{b}_1-a_1{b}_1-a_3{b}_2$

b.

To determine

The area of the red tringle by subtracting the area of the three blue triangles from the area of the rectangle.

Answer to Problem 63E

The area of the red tringle by subtracting the area of the three blue triangles from the area of the rectangle is $\frac{1}{2}\left(a_1{b}_3-a_3{b}_1-a_2{b}_3+a_3{b}_2+a_1{b}_2+a_2{b}_1\right)$

Explanation of Solution

Calculation:

  $\begin{array}{l}B=\frac{1}{2}\left(a_3-a_1|\left(b_1-b_3\right)+\frac{1}{2}\left(a_2-a_3\right)\left(b_3-b_2\right)+\frac{1}{2}\left(a_2-a_1\right)\left(b_2-b_1\right)\right)\\ B=\frac{1}{2}\left(a_1{b}_1+a_3{b}_3-a_1{b}_3-a_3{b}_1-a_2{b}_2-a_3{b}_3+a_2{b}_3+a_3{b}_2+a_2{b}_2+a_1{b}_1-a_1{b}_1-a_2{b}_1\right)\\ B=\frac{1}{2}\left(-a_1{b}_3+b_1{a}_3+a_2{b}_3+a_3{b}_2-a_1{b}_2-a_2{b}_1\right)\end{array}$

So, the area $A$ of the red triangle is then,

  $\begin{array}{l}A=R-B\\ A=\frac{1}{2}\left(a_1{b}_3-a_3{b}_1-a_2{b}_3+a_3{b}_2+a_1{b}_2+a_2{b}_1\right)\end{array}$

Conclusion:

Hence, the area is $\frac{1}{2}\left(a_1{b}_3-a_3{b}_1-a_2{b}_3+a_3{b}_2+a_1{b}_2+a_2{b}_1\right)$ .

c.

To determine

By using $\frac{1}{2}\left(a_1{b}_3-a_3{b}_1-a_2{b}_3+a_3{b}_2+a_1{b}_2+a_2{b}_1\right)$ show that the area of the red triangle is given by

  $area=\pm \frac{1}{2}\left|\begin{array}{ccc}{a}_1& b_1& 1\\ a_2& b_2& 1\\ a_3& b_3& 1\end{array}\right|$

  

Answer to Problem 63E

Showed that the area of the rectangle is giving by the $area=\pm \frac{1}{2}\left|\begin{array}{ccc}{a}_1& b_1& 1\\ a_2& b_2& 1\\ a_3& b_3& 1\end{array}\right|$

Explanation of Solution

Given:

  

  $area=\pm \frac{1}{2}\left|\begin{array}{ccc}{a}_1& b_1& 1\\ a_2& b_2& 1\\ a_3& b_3& 1\end{array}\right|$

Calculation:

First expand the determinant given the 3^rd column,

  $\begin{array}{l}\left|\begin{array}{ccc}{a}_1& b_1& 1\\ a_2& b_2& 1\\ a_3& b_3& 1\end{array}\right|=\left|\begin{array}{cc}{a}_2& b_2\\ a_3& b_3\end{array}\right|-\left|\begin{array}{cc}{a}_1& b_1\\ a_3& b_3\end{array}\right|+\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|\\ =\left(a_2{b}_3-b_2{a}_3\right)-\left(a_1{b}_3-b_1{a}_3\right)+\left(a_1{b}_2-b_1{a}_2\right)\\ =-a_1{b}_3+a_3{b}_1+a_2{b}_3-a_3{b}_2+a_1{b}_2-a_2{b}_1\end{array}$

The sign of the difference arises from the loss of generality because the ordering picked for $a_1,a_2,a_3$ and $b_1,b_2,b_3$ is specific as given in the diagram in the book. So hence write the determinant in an absolute value to fix this and the final result is the proof as,

  $A=\frac{1}{0}abs\left(\left|\begin{array}{ccc}{a}_1& b_1& 1\\ a_2& b_2& 1\\ a_3& b_3& 1\end{array}\right|\right)$

Conclusion:

Hence, area of the rectangle is giving by the $area=\pm \frac{1}{2}\left|\begin{array}{ccc}{a}_1& b_1& 1\\ a_2& b_2& 1\\ a_3& b_3& 1\end{array}\right|$ is proved.