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Chapter 1: Problem 14

Find an equation of the line described. Then sketch the line. The line through \((-3,-2)\) with slope 1

Short Answer

Expert verified
The line's equation is \(y = x + 1\), a line through points like (0,1) and (1,2).

Step by step solution

01

Understand the Information Given

You have a point the line passes through \((-3, -2)\) and a slope of 1. With this information, we can use the point-slope form of the equation of a line.
02

Use the Point-Slope Form

The point-slope form is given by:\[y - y_1 = m(x - x_1)\]where \((x_1, y_1)\) is the point the line passes through and \(m\) is the slope. Substitute the given point and slope: \[y + 2 = 1(x + 3)\]
03

Simplify the Equation

Distribute and simplify the equation from the point-slope form:\[y + 2 = x + 3\]Subtract 2 from both sides:\[y = x + 1\]
04

Identify the Equation Format

The equation \(y = x + 1\) is in the slope-intercept form \(y = mx + b\), which makes it easy to understand that the slope \(m = 1\) and the y-intercept \(b = 1\).
05

Sketch the Line

To sketch the line, plot the y-intercept \((0,1)\), and use the slope to find another point. From \((0,1)\), move 1 unit up and 1 unit right to reach \((1,2)\). Draw the line through these points. Also make sure to check that the line passes through the point \((-3,-2)\).

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Key Concepts

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

Point-Slope Form
The point-slope form is useful when you know the slope of a line and a point through which it passes. It's expressed with the formula:
  • \( y - y_1 = m(x - x_1) \)
In this equation, \( (x_1, y_1) \) represents the coordinates of a specific point on the line, and \( m \) is the slope.
This form helps you find the equation of a line quickly when provided with minimal information. To use it effectively:
  • Recognize the given point, such as \( (-3, -2) \), as \( (x_1, y_1) \).
  • Identify the slope \( m \), which in this case is 1.
Substitute these values into the point-slope form. This allows you to convert specific numerical information into a general equation. Here, we substitute as follows:
  • Point: \( (-3, -2) \), hence \( x_1 = -3 \) and \( y_1 = -2 \)
  • Slope: \( m = 1 \)
Hence, substituting in yields:
  • \( y + 2 = 1(x + 3) \)
This form creates a simple way to build the equation of a line tailored to given conditions.
Slope-Intercept Form
Once you have an equation in point-slope form, you can rearrange it into the slope-intercept form for easier graphing and interpretation. The slope-intercept form is:
  • \( y = mx + b \)
Here, \( m \) is the slope, and \( b \) is the y-intercept, where the line crosses the y-axis.
Starting from the point-slope equation of \( y + 2 = x + 3 \), simplify it into the slope-intercept form:
  • Subtract 2 to isolate \( y \): \( y = x + 1 \)
In the resulting equation \( y = x + 1 \):
  • The slope \( m \) is 1, showing the line rises 1 unit for every 1 unit it runs horizontally.
  • The y-intercept \( b \) is 1, indicating the line crosses the y-axis at the point \( (0, 1) \).
This form not only provides a quick method to graph linear equations but also offers insight into the behavior of the line regarding its steepness and where it intercepts the y-axis.
Graphing Linear Equations
Graphing linear equations involves plotting the equation on a coordinate plane to visualize its path. Here are steps to effectively graph the equation:
  • Begin by identifying the y-intercept, which is where the line crosses the y-axis. From \( y = x + 1 \), the y-intercept is \( 1 \), so plot the point \( (0,1) \).
  • Use the slope to find another point. The slope \( m = 1 \), indicating for every unit you move right (positive direction on the x-axis), you move 1 unit up (positive direction on the y-axis). From \( (0,1) \), move to \( (1,2) \).
Using these two points, \( (0,1) \) and \( (1,2) \), draw a straight line through them with a ruler for accuracy.
Double-check your line by verifying that it passes through any additional given points, such as \( (-3,-2) \), reinforcing the correctness of your plot.
Graphing offers a visual interpretation of the equation and validates calculations by showing the actual trajectory and key characteristics of the line.

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Most popular questions from this chapter

Let \(f(x)=|x-1|+|x+1|\). a. Determine whether the graph of \(f\) is symmetric with respect to either axis or the origin. b. Find alternative expressions for \(f(x)\) in the three cases \(x<-1,-1 \leq x \leq 1\), and \(x>1\), and use this information to sketch the graph of \(f\).

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