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

Answer to Problem 63E
The area of the rectangle is R=(a3−a1).(b2−b1)=a3b2+a1b1−a1b1−a3b2
Explanation of Solution
First finding the coordinates by seeing how the rectangle is formed by the lines x=a1,x=a3,y=b1,y=b2.
So, the line equation by noting he coordinates of the points on them. So, the four vertices of the rectangle are given by,
(a1,b1),(a1,b2),(a3,b1),(a3,b2)
The area of the rectangle id then simply,
R=(a3−a1).(b2−b1)=a3b2+a1b1−a1b1−a3b2
Conclusion:
Hence, the area of the rectangle is R=(a3−a1).(b2−b1)=a3b2+a1b1−a1b1−a3b2
b.
The area of the red tringle by subtracting the area of the three blue triangles from the area of the rectangle.
b.

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 12(a1b3−a3b1−a2b3+a3b2+a1b2+a2b1)
Explanation of Solution
Calculation:
B=12(a3−a1|(b1−b3)+12(a2−a3)(b3−b2)+12(a2−a1)(b2−b1))B=12(a1b1+a3b3−a1b3−a3b1−a2b2−a3b3+a2b3+a3b2+a2b2+a1b1−a1b1−a2b1)B=12(−a1b3+b1a3+a2b3+a3b2−a1b2−a2b1)
So, the area A of the red triangle is then,
A=R−BA=12(a1b3−a3b1−a2b3+a3b2+a1b2+a2b1)
Conclusion:
Hence, the area is 12(a1b3−a3b1−a2b3+a3b2+a1b2+a2b1) .
c.
By using 12(a1b3−a3b1−a2b3+a3b2+a1b2+a2b1) show that the area of the red triangle is given by
area = ±12|a1b11a2b21a3b31|
c.

Answer to Problem 63E
Showed that the area of the rectangle is giving by the area = ±12|a1b11a2b21a3b31|
Explanation of Solution
Given:
area = ±12|a1b11a2b21a3b31|
Calculation:
First expand the determinant given the 3rd column,
|a1b11a2b21a3b31|=|a2b2a3b3|−|a1b1a3b3|+|a1b1a2b2|=(a2b3−b2a3)−(a1b3−b1a3)+(a1b2−b1a2)=−a1b3+a3b1+a2b3−a3b2+a1b2−a2b1
The sign of the difference arises from the loss of generality because the ordering picked for a1,a2,a3 and b1,b2,b3 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=10abs(|a1b11a2b21a3b31|)
Conclusion:
Hence, area of the rectangle is giving by the area = ±12|a1b11a2b21a3b31| is proved.
Chapter 10 Solutions
Precalculus: Mathematics for Calculus - 6th Edition
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