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Chapter 9: Problem 53

Parameterize the equation of the line given in standard form. $$ 2 x+y-3=0 $$

Short Answer

Expert verified
Parameterize by \((x, y) = (t, -2t + 3)\) where \(t\) is a parameter.

Step by step solution

01

Rearrange into Slope-Intercept Form

Start with the given standard form equation: \(2x + y - 3 = 0\). Solve for \(y\) by isolating it on one side of the equation. This is done by subtracting \(2x\) and adding 3 to both sides:\[ y = -2x + 3 \]This is the equation in slope-intercept form \(y = mx + b\), where \(m = -2\) and \(b = 3\).
02

Identify a Parameterization for the Line

To parameterize the line, express \(x\) and \(y\) in terms of a parameter \(t\). The parameterization should satisfy the equation found in the slope-intercept form.Choose \(x = t\). Then substitute \(x = t\) into the slope-intercept form equation:\[ y = -2t + 3 \]Thus, the parameterization of the line is:\[ (x, y) = (t, -2t + 3) \]
03

Verify Parameterization

Substitute the parameterized coordinates back into the original equation to verify it satisfies the equation.Substitute \(x = t\) and \(y = -2t + 3\) into the original equation:\[ 2(t) + (-2t + 3) - 3 = 0 \]Simplify:\[ 2t - 2t + 3 - 3 = 0 \]\[ 0 = 0 \]This confirms the parameterization is correct.

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Key Concepts

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

Understanding Standard Form
The standard form of a linear equation is expressed as \(Ax + By = C\). In this form, \(A\), \(B\), and \(C\) are constants.
The main advantage of using the standard form is that it provides a quick way to identify the intercepts of a line.
  • **Intercepts:** You can easily find the x-intercept by setting \(y = 0\) and solving for \(x\). Similarly, find the y-intercept by setting \(x = 0\) and solving for \(y\).
  • **Flexibility:** Standard form is flexible and can accommodate more complex systems of equations.

However, directly graphing from this form can be less intuitive, which is why equations are often converted into slope-intercept form for better visualization.
Revealing Slope-Intercept Form
The slope-intercept form of a line is expressed as \(y = mx + b\). Here, \(m\) represents the slope and \(b\) is the y-intercept of the line.
This form is preferred when you are interested in quickly identifying and graphing a line because:
  • **Slope:** The slope \(m\) indicates the steepness and direction of the line. A positive slope means the line rises, while a negative slope means it falls.
  • **Y-Intercept:** The y-intercept \(b\) tells you where the line crosses the y-axis, making it easy to start graphing.

In the original problem, converting from standard form \(2x + y - 3 = 0\) to slope-intercept form yields \(y = -2x + 3\). This clearly shows a slope of -2 and a y-intercept of 3, assisting with visualizing the line's behavior.
Exploring Line Parameterization
Line parameterization involves expressing the coordinates \((x, y)\) of a line in terms of a single parameter, usually denoted as \(t\). This approach not only provides a dynamic way to describe a line but is also useful in mathematical computations.
When parameterizing a line:
  • **Flexibility:** Choose one variable to express in terms of the parameter. In our problem, we let \(x = t\). This choice simplifies the process.
  • **Dependence:** Substitute the parameterized form into the slope-intercept equation to find the corresponding \(y\) value, e.g., \(y = -2t + 3\).

Thus, the parameterized equation becomes \((x, y) = (t, -2t + 3)\), where \(t\) can take any real value. This perspective helps in understanding lines as sets of points continuously connected along the parameter \(t\).

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Most popular questions from this chapter

Write each system in matrix form. $$ \begin{array}{r} 2 x_{2}-x_{1}=x_{3} \\ 4 x_{1}+x_{3}=7 x_{2} \\ x_{2}-x_{1}=x_{3} \end{array} $$

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Show that } A+B=B+A \text { . } $$

Reduce the system of linear equations to upper triangular form and solve. $$ \begin{array}{l} 7 x-y=4 \\ 3 x+2 y=1 \end{array} $$

Let $$ A=\left[\begin{array}{rr} 1 & 3 \\ 0 & -2 \end{array}\right] \text { and } I_{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $$ Show that \(A I_{2}=I_{2} A=A\).

Solve each system of linear equations. $$ \begin{array}{l} 2 x-y+3 z=3 \\ 2 x+y+4 z=4 \\ 2 x-3 y+2 z=2 \end{array} $$
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