Problem 23
Question
Determine the equation of the line that satisfies the stated requirements. Put the equation in standard form. The line with slope \(-2\) and \(x\) -intercept \((1,0)\)
Step-by-Step Solution
Verified Answer
The equation of the line in standard form is \(2x + y = 2\).
1Step 1: Understand the Problem
The problem asks for the equation of a line with a given slope and an x-intercept. Specifically, the slope is \(-2\) and the x-intercept is \((1,0)\). The equation should be in standard form \(Ax + By = C\).
2Step 2: Use the Point-Slope Form
Using the point-slope formula, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (1, 0)\) and \(m = -2\), substitute these values to get: \[ y - 0 = -2(x - 1) \] which simplifies to: \[ y = -2x + 2 \]
3Step 3: Convert to Slope-Intercept Form
The equation from the previous step \(y = -2x + 2\) is in slope-intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
4Step 4: Rearrange to Standard Form
Convert the slope-intercept form \(y = -2x + 2\) to standard form \(Ax + By = C\) by rearranging terms: Add \(2x\) to both sides: \[ 2x + y = 2 \]This is the standard form of the equation.
Key Concepts
Slope-Intercept FormStandard FormPoint-Slope Form
Slope-Intercept Form
The slope-intercept form of a linear equation is a popular way to express a line and is written as \( y = mx + b \). Here, the letter \( m \) stands for the slope of the line. The slope indicates how steep the line is. A positive slope means the line ascends as it moves from left to right, while a negative slope means it descends. The letter \( b \) is the y-intercept, the point where the line crosses the y-axis. This form is particularly useful when you need to quickly identify both the slope and the y-intercept.
The slope-intercept form makes it easy to graph linear equations because you can start at the y-intercept \( (0, b) \) on the graph and use the slope \( m \) to determine another point. For instance, a slope of \( -2 \) means that for every one unit you move to the right along the x-axis, you go down two units on the y-axis.
The slope-intercept form makes it easy to graph linear equations because you can start at the y-intercept \( (0, b) \) on the graph and use the slope \( m \) to determine another point. For instance, a slope of \( -2 \) means that for every one unit you move to the right along the x-axis, you go down two units on the y-axis.
Standard Form
The standard form of a linear equation is expressed as \( Ax + By = C \), where \( A \), \( B \), and \( C \) are integers, and \( A \) should ideally be a non-negative number. This form is advantageous when you need to analyze the line in relation to both the x and y axes or when dealing with systems of equations.
To convert from slope-intercept form to standard form, as shown in the original problem, you'll rearrange the equation to get all variable terms on one side:
To convert from slope-intercept form to standard form, as shown in the original problem, you'll rearrange the equation to get all variable terms on one side:
- Start with the slope-intercept form: \( y = -2x + 2 \).
- Add \( 2x \) to both sides to eliminate the x-term from the right side: \( 2x + y = 2 \).
Point-Slope Form
Point-slope form is written as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope, exactly like in the original problem. When you know a specific point on a line and the line's slope, this form is very handy.
The point-slope form is particularly useful when you're given or already know a point and the slope, because you can quickly construct the line's equation. In the sample problem:
The point-slope form is particularly useful when you're given or already know a point and the slope, because you can quickly construct the line's equation. In the sample problem:
- Take the point \( (1,0) \) and slope \( -2 \).
- Substitute into the formula: \( y - 0 = -2(x - 1) \).
Other exercises in this chapter
Problem 23
Explain how the following functions can be obtained from \(y=x^{2}\) by basic transformations: (a) \(y=x^{2}-2\) (b) \(y=(x-1)^{2}+1\) (c) \(y=-2(x+2)^{2}\)
View solution Problem 23
Graph \(y=x^{n}, x \geq 0\), for \(n=1,2,3\), and 4 in one coordinate system. Where do the curves intersect?
View solution Problem 24
Explain how the following functions can be obtained from \(y=x^{3}\) by basic transformations: (a) \(y=x^{3}-1\) (b) \(y=-x^{3}-1\) (c) \(y=-3(x-1)^{3}\)
View solution Problem 24
(a) Graph \(f(x)=x, x \geq 0\), and \(g(x)=x^{2}, x \geq 0\), together, in one coordinate system. (b) For which values of \(x\) is \(f(x) \geq g(x)\), and for w
View solution