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Chapter 1: Problem 12

Determine the equation of the line that satisfies the stated requirements. Put the equation in standard form. The line passing through \((-1,4)\) and \(\left(2,-\frac{1}{2}\right)\)

Short Answer

Expert verified
The equation is \(3x + 2y = 5\).

Step by step solution

01

Calculate the slope of the line

We begin by calculating the slope \( m \) of the line using the formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Plug in the coordinates: \( (x_1, y_1) = (-1, 4) \) and \( (x_2, y_2) = \, \left(2, -\frac{1}{2}\right) \). This gives us: \[ m = \frac{-\frac{1}{2} - 4}{2 + 1} = \frac{-\frac{1}{2} - \frac{8}{2}}{3} = \frac{-\frac{9}{2}}{3} = -\frac{3}{2} \]. The slope of the line is \(-\frac{3}{2}\).
02

Use the point-slope form

Utilize the point-slope form of the line equation: \( y - y_1 = m(x - x_1) \). Choose the point \((-1, 4)\) and substitute \(m = -\frac{3}{2}\): \[ y - 4 = -\frac{3}{2}(x + 1) \].
03

Simplify to slope-intercept form

Distribute the slope on the right-hand side: \[ y - 4 = -\frac{3}{2}x - \frac{3}{2} \]. Add 4 to both sides to isolate \(y\): \[ y = -\frac{3}{2}x - \frac{3}{2} + 4 \]. Simplify \( \frac{8}{2} = 4 \) to get \( y = -\frac{3}{2}x + \frac{5}{2} \).
04

Convert to standard form

Multiply the entire equation by 2 to eliminate the fractions: \[ 2y = -3x + 5 \]. Rearrange terms to get the standard form: \( Ax + By = C \). This becomes \[ 3x + 2y = 5 \]. The equation in standard form is \( 3x + 2y = 5 \).

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Key Concepts

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

Slope Calculation
To determine the slope of a line passing through two points, we use the formula for slope, given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \). This formula helps us understand how steep a line is by calculating the ratio of the vertical change to the horizontal change between two points on the line. Let's break it down:
  • Vertical Change (Rise): This is the difference in the y-values of the two points. In our example, it is \(-\frac{1}{2} - 4\), simplifying to \(-\frac{9}{2}\).
  • Horizontal Change (Run): This is the difference in the x-values. For our points \((-1, 4)\) and \( \left(2, -\frac{1}{2}\right)\), the horizontal change is \(2 + 1 = 3\).
Putting it together, the slope is \(-\frac{9}{2} \div 3 = -\frac{3}{2}\). This tells us the line decreases by \(1.5\) units for each unit it moves to the right.
Standard Form Equation
The standard form of a linear equation is expressed as \( Ax + By = C \), where \( A \), \( B \), and \( C \) are integers, and \( A \) should be a non-negative number. This form is useful for identifying key properties of the line and is often required in algebraic problems.
  • Why Standard Form? This form makes it straightforward to find the x-intercept and y-intercept, understand the relationship of x and y, and easily compare different lines.
  • Steps to Achieve Standard Form: After finding the slope, and expressing the line in slope-intercept form \( y = mx + b \), the next task is to convert it into standard form. This involves getting rid of fractions through multiplication and rearranging the equation.
In our example, starting with \( y = -\frac{3}{2}x + \frac{5}{2} \), we multiply every term by 2 to remove fractions, resulting in \( 2y = -3x + 5 \). Reordering gives us \( 3x + 2y = 5 \), which is the standard form of the equation.
Point-Slope Form
The point-slope form of an equation is particularly advantageous when we're given a point on the line and the line's slope. It is written as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a known point, and \( m \) is the slope.
  • Practical Use: Quickly find an equation of the line when you know one point and the slope. It's handy for quickly converting information into a usable equation format.
  • Substitution Example: Using the point \((-1, 4)\) and slope \(-\frac{3}{2}\), we have: \( y - 4 = -\frac{3}{2}(x + 1) \). This form is clear and easy to use for further simplifications or transformations, such as getting to slope-intercept form or standard form.
By leveraging the point-slope form, we simplify the process of defining a line, making it a vital tool in algebraic methods.

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