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Chapter 8: Problem 25

Finding the Area of a Triangle In Exercises \(17-28\) , use a determinant to find the area with the given vertices. $$(-3,5),(2,6),(3,-5)$$

Short Answer

Expert verified
The area of the triangle is 28 square units.

Step by step solution

01

Determine the formula and insert the coordinates

Use the formula for the area of a triangle with the given vertices. Substitute \((-3,5),(2,6),(3,-5)\) into the formula: \[A = 0.5 * |-3 * (6 - (-5)) + 2 * (-5 - 5) + 3 * (5 - 6)|)\]
02

Simplify

Simplify the equation to: \[A = 0.5 * |-3 * 11 + 2 * -10 + 3 * -1|\]
03

Calculate the area

Calculate the result from: \[ A = 0.5 * (-33 - 20 - 3)\]
04

Final calculation

Further simplify the expression: \[A = 0.5 * -56 = -28\]
05

Take absolute value

As area cannot be negative, let's take the absolute value of the result. Thus, \[A = |-28| = 28\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinants
Determinants are mathematical tools that can help find the area of geometric shapes, like triangles, using algebraic expressions. They are particularly useful when working with matrices. The formula for the area of a triangle using determinants involves placing the coordinates of the vertices into a matrix and applying the determinant formula.
  • For three vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), the determinant helps calculate:\[A = 0.5 \times \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|\]
  • This formula captures the "signed area," hence the need to take the absolute value at the end to ensure the area is positive, as seen in the exercise.
Determinants are effective as they allow computation without having to rearrange geometric expressions, streamlining the process.
Coordinate Geometry
Coordinate geometry, or analytic geometry, is the field of mathematics where algebraic equations are used to describe geometric concepts. This approach allows us to translate geometric problems, like finding the area, into mathematical language.
  • In coordinate geometry, points are represented as ordered pairs \((x, y)\), with the x-axis and y-axis providing reference.
  • Using this framework, distances, slopes, and areas between points can be calculated systematically.
  • Particularly for triangles, coordinate geometry simplifies calculating areas by placing vertices within a coordinate plane and applying algebraic formulas like the determinant method.
Understanding how points relate within a coordinate system is crucial for tackling problems involving shapes and their properties.
Vertices of a Triangle
Vertices are the corner points of a triangle and are crucial in determining its properties, such as area and perimeter. In a triangle with vertices \((A, B, C)\), the coordinates of these points allow for the application of various mathematical techniques.
  • Each vertex of a triangle is an intersection of two of its sides and collectively, the vertices determine the shape's positioning in space.
  • To find the area using the determinant method, knowing the exact coordinates of the vertices is essential.
  • The coordinates are substituted directly into the formula, as seen in the exercise with vertices \((-3,5), (2,6), (3,-5)\).
By knowing the vertices, we can create the matrix needed for the determinant calculation, simplifying the problem of finding the triangle's area using algebraic operations.

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Most popular questions from this chapter

Properties of Determinants In Exercises \(101-103\) ,a property of determinants is given \((A\) and \(B\) are squarematrices). State how the property has been applied to the given determinants and use a graphing utility to verify the results. Properties of Determinants In Exercises \(101-103\) ,a property of determinants is given \((A\) and \(B\) are squarematrices). State how the property has been applied to the given determinants and use a graphing utility to verify the results. If \(B\) is obtained from \(A\) by interchanging two rows of \(A\) or interchanging two columns of \(A,\) then \(|B|=-|A|\) $$(a)\left| \begin{array}{rrr}{1} & {3} & {4} \\ {-7} & {2} & {-5} \\ {6} & {1} & {2}\end{array}\right|=-\left| \begin{array}{rrr}{1} & {4} & {3} \\ {-7} & {-5} & {2} \\ {6} & {2} & {1}\end{array}\right|$$ $$(b)\left| \begin{array}{r|rrr}{1} & {3} & {4} \\ {-2} & {2} & {0} \\ {1} & {6} & {2}\end{array}\right|=-\left| \begin{array}{rrr}{1} & {6} & {2} \\ {-2} & {2} & {0} \\ {1} & {3} & {4}\end{array}\right|$$

Comparing Linear Systems and Matrix Operations In Exercises 41 and \(42,\) (a) perform the row operations to solve the augmented matrix, (b) write and solve the system of linear equations represented by the augmented matrix, and (c) compare the two solution methods. Which do you prefer? $$\left[ \begin{array}{rrrr}{-3} & {4} & {\vdots} & {22} \\ {6} & {-4} & {\vdots} & {-28}\end{array}\right]$$ $$\begin{array}{l}{\text { (i) Add } R_{2} \text { to } R_{1} \text { . }} \\\ {\text { (ii) Add }-2 \text { times } R_{1} \text { to } R_{2} \text { . }} \\\ {\text { (iii) Multiply } R_{2} \text { by }-\frac{1}{4}} \\ {\text { (iv) Multiply } R_{1} \text { by } \frac{1}{3}}\end{array}$$

Gaussian Elimination with Back-Substitution, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution. $$\left\\{\begin{aligned} 3 x+2 y-z+w &=0 \\ x-y+4 z+2 w &=25 \\\\-2 x+y+2 z-w &=2 \\ x+y+z+w &=6 \end{aligned}\right.$$

Using a Graphing Utility, use the matrix capabilities of a graphing utility to write the augmented matrix corresponding to the system of equations in reduced row-echelon form. Then solve the system. $$\left\\{\begin{aligned} x+2 y+2 z+4 w=& 11 \\ 3 x+6 y+5 z+12 w=& 30 \\ x+3 y-3 z+2 w=&-5 \\ 6 x-y-z+\quad w=&-9 \end{aligned}\right.$$

Row-Echelon Form, determine whether the matrix is in row-echelon form. If it is, determine whether it is in reduced row-echelon form. $$\left[ \begin{array}{llll}{1} & {3} & {0} & {0} \\ {0} & {0} & {1} & {8} \\\ {0} & {0} & {0} & {0}\end{array}\right]$$
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