Precalculus: Mathematics for Calculus - 6th Edition
Precalculus: Mathematics for Calculus - 6th Edition
6th Edition
ISBN: 9780840068071
Author: Stewart, James, Redlin, Lothar, Watson, Saleem
Publisher: Cengage Learning
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Question
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Chapter 10.6, Problem 63E

a.

To determine

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

a.

Expert Solution
Check Mark

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.

To determine

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

b.

Expert Solution
Check Mark

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.

To determine

By using 12(a1b3−a3b1−a2b3+a3b2+a1b2+a2b1) show that the area of the red triangle is given by

  area = ±12|a1b11a2b21a3b31|

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 10.6, Problem 63E , additional homework tip 1

c.

Expert Solution
Check Mark

Answer to Problem 63E

Showed that the area of the rectangle is giving by the area = ±12|a1b11a2b21a3b31|

Explanation of Solution

Given:

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 10.6, Problem 63E , additional homework tip 2

  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.

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Chapter 10 Solutions

Precalculus: Mathematics for Calculus - 6th Edition

Ch. 10.1 - The system of equations {2x+3y=75xy=9 is a system...Ch. 10.1 - Prob. 2ECh. 10.1 - A system of two linear equations in two variables...Ch. 10.1 - The following is a system of two linear equations...Ch. 10.1 - Prob. 5ECh. 10.1 - Prob. 6ECh. 10.1 - Prob. 7ECh. 10.1 - Prob. 8ECh. 10.1 - Prob. 9ECh. 10.1 - Prob. 10E
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An Introduction to Area | Teaching Maths | EasyTeaching; Author: EasyTeaching;https://www.youtube.com/watch?v=_uKKl8R1xBM;License: Standard YouTube License, CC-BY
Area of a Rectangle, Triangle, Circle & Sector, Trapezoid, Square, Parallelogram, Rhombus, Geometry; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=JnLDmw3bbuw;License: Standard YouTube License, CC-BY