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Chapter 9: Problem 12

In Exercises \(7-24,\) find the general form of the equation of the line satisfying the conditions given and graph the line. Through \((-7,1)\) and \((3,-5)\)

Short Answer

Expert verified
The general form of the equation is \( 3x + 5y + 16 = 0 \).

Step by step solution

01

- Determine the Slope

First, calculate the slope using the formula \( m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \). Substituting the given points \( (-7, 1) \) and \( (3, -5) \): \ m = \frac{{-5 - 1}}{{3 - (-7)}} = \frac{{-6}}{{10}} = -\frac{3}{5} \.
02

- Use Point-Slope Form

Using the point-slope form \( y - y_1 = m(x - x_1) \), substitute \( (-7, 1) \) and the calculated slope \( m = -\frac{3}{5} \): \ y - 1 = -\frac{3}{5}(x + 7) \.
03

- Simplify to Slope-Intercept Form

Simplify the equation to put it in the slope-intercept form \( y = mx + b \): \ y - 1 = -\frac{3}{5}(x + 7) \ \ y - 1 = -\frac{3}{5}x - \frac{21}{5} \ \ y = -\frac{3}{5}x - \frac{21}{5} + 1 \ \ y = -\frac{3}{5}x - \frac{21}{5} + \frac{5}{5} \ \ y = -\frac{3}{5}x - \frac{16}{5} \.
04

- Convert to General Form

Rewrite the equation in general form \( Ax + By + C = 0 \): \ y = -\frac{3}{5}x - \frac{16}{5} \ \ 5y = -3x - 16 \ \ 3x + 5y + 16 = 0 \.
05

- Graph the Line

Plot the points \( (-7, 1) \) and \( (3, -5) \) on graph paper. Draw the line passing through these points to visualize the equation \( 3x + 5y + 16 = 0 \).

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Key Concepts

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

slope calculation
Calculating the slope is the first step in finding the equation of a line. The slope tells us how steep the line is. For any two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the slope \( m \) is given by the formula \( m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \). This relationship measures the vertical change (rise) divided by the horizontal change (run) between the two points.

Let's apply this to our points \( (-7, 1) \) and \( (3, -5) \):
\ m = \frac{{-5 - 1}}{{3 - (-7)}} = \frac{{-6}}{{10}} = -\frac{3}{5} \

This negative slope indicates that the line falls as it moves from left to right.
point-slope form
Point-slope form makes use of the slope and one point on the line. The formula is \( y - y_1 = m(x - x_1) \). This is useful when we already know the slope and at least one point through which the line passes.

Using the calculated slope \( m = -\frac{3}{5} \) and point \( (-7, 1) \):

\ y - 1 = -\frac{3}{5}(x + 7) \

We use this form to quickly plug in the known values without needing to rearrange the equation. This form is especially useful when working out the next step: converting to slope-intercept form.
slope-intercept form
The slope-intercept form is \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept (where the line crosses the y-axis). This form is very common in algebra because it clearly shows the slope and the y-intercept.

Starting from point-slope form \( y - 1 = -\frac{3}{5}(x + 7) \), let's simplify:

\begin{align*} y - 1 &= -\frac{3}{5}x - \frac{21}{5} \ y &= -\frac{3}{5}x - \frac{21}{5} + 1 \ y &= -\frac{3}{5}x - \frac{21}{5} + \frac{5}{5} \ y &= -\frac{3}{5}x - \frac{16}{5} \end{align*}

This gives us the slope-intercept form: \( y = -\frac{3}{5}x - \frac{16}{5} \).
general form
The general form of a linear equation is \( Ax + By + C = 0 \), where \( A \), \( B \), and \( C \) are integers, and \( A \) should be a non-negative integer. This form is particularly useful for solving systems of linear equations and graphing.

From slope-intercept form \( y = -\frac{3}{5}x - \frac{16}{5} \), we need to eliminate fractions and rearrange terms:
\ 5y = -3x - 16 \
\ 3x + 5y + 16 = 0 \

Now we have the general form: \( 3x + 5y + 16 = 0 \). Notice how every coefficient is an integer. This equation can easily be used for various applications, including graphing and analyzing linear systems.

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Most popular questions from this chapter

Suppose the origin is the midpoint of a segment and one endpoint of the segment is \((2,-4) .\) Find the coordinates of the other endpoint.

In Exercises \(1-6,\) find the slope of the line passing through the given pair of points. \((-5,2)\) and \((-1,-6)\)

In Exercises \(33-38\), determine whether the lines with the given equations are parallel, perpendicular, or neither. \(2 x-y+7=0\) \(-6 x+3 y-1=0\)

In Exercises \(43-46\), find the general form of the equation of the line perpendicular to the line that contains the given points and that passes through the point midway between them. \((-5,1)\) and \((3,-1)\)

Plot the points associated with the ordered pairs \(P\left(-\frac{1}{2}, 3\right), Q\left(\frac{7}{2}, 4\right), R\left(-3,-\frac{3}{4}\right),\) and \(S\left(2,-\frac{7}{3}\right)\)
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