8.2 Graphing Linear Equations

A linear equation is an algebraic equation with one or two variables (each with an exponent of one), which produces a straight line when plotted on a graph.

Examples of linear equations with one variable in the rectangular coordinate system are:

[latex] 3x - 5 = 0 [/latex],   [latex] x = 3 [/latex],   [latex] y + 2 = 0 [/latex],   [latex] \displaystyle{y = -\frac{7}{5}} [/latex]

The process of finding the value of the variables for which the equation is true is known as solving the equation. A linear equation with two variables has infinitely many pairs of values as solutions; therefore, it is often convenient to represent these solutions by drawing a graph.

Linear equations with two variables are generally represented by the variables x and y and are expressed either in the form of Ax + By = C (where A, B, and C are integers and AA is positive), known as standard form, or in the form of y = mx + b (where m and b are integers or fractions), known as slope-intercept form.

Linear Equations in Standard Form

The 'standard' form for a linear equation with two variables, x and y, is written as Ax + By = C, where A, B, and C are integers, v is positive, and A, B, and C, have no common factors other than 1.

For example, consider the following simple linear equation with two variables: [latex] 2x - y = -3 [/latex]

This equation is in the standard form of Ax + By = C, where A = 2, B = −1, and C = −3.

• If a given equation has fractions, then multiply each term by the least common denominator (LCD), divide each term by any common factors, and rearrange the equation into standard form.

• If a given equation has decimals, then multiply each term by an appropriate power of 10 to eliminate the decimals, divide each term by any common factors, and rearrange the equation into standard form.

• If a given equation has no fractions or decimals, then divide each term by any common factors and rearrange it into standard form.

Example 8.2-a: Writing Linear Equations in Standard Form

Write the following linear equations in standard form:

1. [latex] \displaystyle{\frac{2}{3}x + \frac{1}{2}y - 3 = 0} [/latex]

   
   

2. [latex] 0.3x = 1.25y - 2 [/latex]

   
   

3. [latex] \displaystyle{y = \frac{2}{3}x - 5} [/latex]

   
   

Solution

1. [latex] \displaystyle{\frac{2}{3}x + \frac{1}{2}y - 3 = 0} [/latex]   Multiplying each term by the LCD of 6 and simplifying,

   [latex] 4x + 3y - 18 = 0 [/latex]   Rearranging,

   [latex] 4x + 3y = 18 [/latex]   This is in the form Ax + By = C.

   Therefore, the equation [latex] \displaystyle{\frac{2}{3}x + \frac{1}{2}y - 3 = 0} [/latex], in standard form, is [latex] 4x + 3y = 18 [/latex]

   
   

2. [latex] 0.3x = 1.25y - 2 [/latex]   Multiplying each term by 100,

   [latex] 30x = 125y - 200 [/latex]   Dividing each term by the common factor 5 and simplifying,

   [latex] 6x = 25y - 40 [/latex]   Rearranging,

   [latex] 6x - 25y = -40 [/latex]   This is in the form Ax + By = C.

   Therefore, the equation [latex] 0.3x = 1.25y - 2 [/latex], in standard form, is [latex] 6x - 25y = -40 [/latex]

   
   

3. [latex] \displaystyle{y = \frac{2}{3}x - 5} [/latex]   Multiplying each term by 3 and simplifying,

   [latex] 3y = 2x - 15 [/latex]   Rearranging,

   [latex] -2x + 3y = -15 [/latex]   Multiplying each term by –1 to make A positive,

   [latex] 2x - 3y = 15 [/latex]   This is in the form Ax + By = C.

   Therefore, the equation [latex] \displaystyle{y = \frac{2}{3}x - 5} [/latex], in standard form, is [latex] 2x - 3y = 15 [/latex]

   
   

Linear Equations in Slope-Intercept Form

The 'slope-intercept' form for a linear equation with two variables, x and y, is written as y = mx + b, where m and b are either integers or fractions. 'm' represents the slope and 'b' represents the y-coordinate of the y-intercept.

The slope is the steepness of the line relative to the X-axis.

The y-intercept is the point at which the line crosses the Y-axis and where the x-coordinate is zero.

For example, consider a simple linear equation such as y = 2x + 3. This equation is in the slopeintercept form of y = mx + b, where m = 2 and b = 3; hence, the slope is 2 and the y-intercept is (0, 3).

Example 8.2-b: Writing Linear Equations in Slope-Intercept Form

Write the following linear equations in slope-intercept form and identify the slope and the y-intercept.

1. [latex] 4x + 3y = 18 [/latex]

   
   

2. [latex] \displaystyle{x = \frac{25}{6}y + \frac{20}{3}} [/latex]

   
   

Solution

1. [latex] 4x + 3y = 18 [/latex]   Rearranging the equation with y on the left,

   [latex] 3y = -4x + 18 [/latex]   Dividing each term by 3 and simplifying,

   [latex] \displaystyle{y = -\frac{4}{3}x + 6} [/latex]   This is in the form y = mx + b.

   Therefore, the slope is [latex] \displaystyle{m = -\frac{4}{3}} [/latex] and the y-intercept is the point (0, 6).

   
   

2. [latex] \displaystyle{x = \frac{25}{6}y + \frac{20}{3}} [/latex]   Multiplying each term by the LCD of 6 and simplifying,

   [latex] 6x = 25y + 40 [/latex]   Rearranging the equation with y on the left,

   [latex] -25y = -6x + 40 [/latex]   Multiplying each term by −1,

   [latex] 25y = 6x - 40 [/latex]   Dividing each term by 25 and simplifying,

   [latex] \displaystyle{y = \frac{6}{25}x - \frac{8}{5}} [/latex]   This is in the form y = mx + b.

   Therefore, the slope is [latex] \displaystyle{m = -\frac{6}{25}} [/latex] and the y-intercept is the point [latex] \displaystyle{(0, -\frac{8}{5})} [/latex].

   
   

Determining the Solution Set of a Linear Equation

If a linear equation has two variables, x and y, then there are infinitely many solutions to the equation, and it is not possible to solve the equation for a single value of each variable. However, it is possible to create a set of solutions by replacing one variable (either x or y) with any value and then computing for the value of the other variable.

For example, consider the equation: [latex] y = 2x + 3 [/latex]

Choosing [latex] x = 1 [/latex] and substituting [latex] x = 1 [/latex] into the equation:

[latex] y = 2x + 3 [/latex]

[latex] y = 2(1) + 3 = 5 [/latex]

Therefore, [latex] x = 1 [/latex] and [latex] y = 5 [/latex] is one of the infinitely many solutions to the equation; i.e., the ordered pair (1, 5) is a point that satisfies the equation.

Choosing [latex] x = 2 [/latex] and substituting [latex] x = 2 [/latex] into the equation:

[latex] y = 2x + 3 [/latex]

[latex] y = 2(2) + 3 = 7 [/latex]

Therefore, [latex] x = 2 [/latex] and [latex] y = 7 [/latex] is another solution to the equation; i.e., the ordered pair (2, 7) is another point that satisfies the equation.

Similarly, we can obtain any number of points that satisfy the equation by choosing different values for x, and computing the corresponding value for y.

We can represent the full solution set by graphing the linear equation. The graph of the linear equation will be a line formed by all the solutions to the linear equation. Conversely, any point on the line is a solution to the linear equation.

Graphing Linear Equations Using a Table of Values

Follow these steps to graph a linear equation using a table of values:

Step 1:   Create a table of values by choosing any value for the variable x (0 is often a good first choice).

Step 2:   Compute the corresponding value for the variable y (this is easiest if the equation is in slope-intercept form).

Step 3:   Form the ordered pair (x, y).

Step 4:   Repeat Steps 1 to 3 at least two more times to create at least three ordered pairs.

Step 5:   Plot the ordered pairs (points) on the coordinate system, using an appropriate scale.

Step 6:   Join the points in a straight line, continuing the line indefinitely in both directions using arrows.

Step 7:   Label the graph with the equation of the line.

For example, consider the linear equation [latex] y = 2x + 3 [/latex]. We will first determine the coordinates of four ordered pairs that are on this line by choosing values for x and finding the corresponding values for y; then we will draw the graph.

• Choosing [latex] x = 0 [/latex],

  [latex] y = 2x + 3 = 2(0) + 3 = 0 + 3 = 3 [/latex],

  (0, 3) is a point on the line.

  
  

• Choosing [latex] x = 1 [/latex],

  [latex] y = 2x + 3 = 2(1) + 3 = 2 + 3 = 5 [/latex],

  (1, 5) is a point on the line.

  
  

• Choosing [latex] x = 2 [/latex],

  [latex] y = 2x + 3 = 2(2) + 3 = 4 + 3 = 7 [/latex],

  (2, 7) is a point on the line.

  
  

• Choosing [latex] x = 3 [/latex],

  [latex] y = 2x + 3 = 2(3) + 3 = 6 + 3 = 9 [/latex],

  (3, 9) is a point on the line.

  
  

Since all the points fall on a line when joined, it verifies that the plotted line represents the equation.

Graphing Linear Equations Using the x-intercept and the y-intercept

Recall that the y-intercept is the point at which the line crosses the Y-axis and where the x-coordinate is zero. Similarly, the x-intercept is the point at which the line crosses the X-axis and where the y-coordinate is zero.

If the intercepts are not at the origin (0, 0), we may use the x-intercept and y-intercept as two points to draw a linear graph and use a third point to test the drawn line.

If the intercepts are at the origin (0, 0), then we need to compute another ordered pair to use as the second point (along with the origin) with which to draw the line, and a third point to test the drawn line.

For example, consider a linear equation, [latex] 3x - y = -9 [/latex], where we will find the x-intercept, y-intercept, and another ordered pair to draw the graph.

Finding the x-intercept:

• Substituting [latex] y = 0 [/latex] into the given equation and solving for x,

• [latex] 3x - 0 = -9 [/latex], thus, [latex] x = -3 [/latex]. Therefore, (−3, 0) is the x-intercept.

Finding the y-intercept:

• Substituting [latex] x = 0 [/latex] into the given equation and solving for y,

• [latex] 3(0) - y = -9 [/latex], thus, [latex] y = 9 [/latex]. Therefore, (0, 9) is the y-intercept.

Finally, finding another ordered pair as a third point on the line to use as a test point to verify the plotted line joining the x-intercept and y-intercept:

• Choosing [latex] x = -1 [/latex], substituting this in the given equation, and solving for y,

• [latex] 3(-1) - y = -9 [/latex], thus, [latex] y = 6 [/latex]. Therefore, (–1, 6) is a point on the line [latex] 3x - y = -9 [/latex].

Now that we have the x-intercept, the y-intercept, and a test point:

• Plot the ordered pairs on the coordinate system using an appropriate scale.

• Draw a line to join the x-intercept and y-intercept.

• Verify that the test point falls on the graph of the plotted line.

Since the point (–1, 6) falls on the line when plotted on the graph, it verifies that the plotted line represents the linear equation.

Graphing Linear Equations Using the Slope and the y-intercept

Recall that a linear equation in the form of [latex] y = mx + b [/latex] is known as the equation in slope-intercept form, where 'm' is the slope and 'b' is the y-coordinate of the y-intercept.

If the equation is in the standard form [latex] Ax + By = C [/latex], it can be rearranged to represent the slope-intercept form, as follows:

[latex] Ax + By = C [/latex]

[latex] By = -Ax + C [/latex]

[latex] \displaystyle{y = -\frac{A}{B}x + \frac{C}{B}} [/latex]

This is in the form [latex] y = mx + b [/latex], where the slope [latex] \displaystyle{m = -\frac{A}{B}} [/latex], and the y-coordinate of the y-intercept [latex] \displaystyle{b = \frac{C}{B}} [/latex].

The Slope and y-intercept of a Line

The slope (m) is the steepness of the line relative to the X-axis. It is the ratio of the change in value of y (called the 'rise' and denoted [latex] \Delta y [/latex]) to the corresponding change in value of x (called the 'run' and denoted [latex] \Delta x [/latex]).

If P ([latex] x_1, y_1 [/latex]) and Q ([latex] x_2, y_2 [/latex]) are two different points on a line, then the slope of the line PQ between the two points is:

[latex] \boldsymbol{\displaystyle{m = \frac{Rise}{Run} = \frac{Change in y value}{Change in x value} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}}} [/latex]

This is illustrated in Exhibits 8.2-c and 8.2-d:

Example 8.2-c: Finding the Slope and y-intercept of a Linear Equation and Graphing the Equation

Find the slope and y-intercept of the linear equation [latex] -2x + 3y - 12 = 0 [/latex] and graph the equation.

Solution

[latex] 4x + 3y = 18 [/latex]   Rearranging the equation with y on the left,

[latex] 3y = -4x + 18 [/latex]   Dividing each term by 3 and simplifying,

[latex] \displaystyle{y = -\frac{4}{3}x + 6} [/latex]   This is in the form y = mx + b.

Therefore, the slope is [latex] \displaystyle{m = -\frac{4}{3}} [/latex] and the y-intercept is the point (0, 6).

Therefore, (0, 4) is a point on the line and the slope, [latex] \displaystyle{m = \frac{Rise}{Run} = \frac{Change in y value}{Change in x value} = \frac{2}{3}} [/latex], represents an increase of 2 in the vertical direction ('rise') for every increase of 3 in the horizontal direction ('run').

Representing this on a graph:

1. First, plot the y-intercept (0, 4).

2. From this point, move 3 units to the right and then move 2 units up to locate the new point (3, 6).
   Note: This is the same as moving 2 units up, and then 3 units to the right.

3. Similarly, from the point (3, 6), move 3 units to the right and 2 units up to locate another point (6, 8).

4. Draw the line through these points to graph the equation.

   
   

Or,

1. First, plot the y-intercept (0, 4).

2. From this point, move 3 units to the left and then move 2 units down to locate the new point (−3, 2).

3. Similarly, from the point (−3, 2), move 3 units to the left and 2 units down to locate another point (−6, 0).

4. Draw the line through these points to graph the equation.

   
   

Note: All the points must lie on the same line or else a mistake has been made.

Example 8.2-d: Graphing a Linear Equation in the Slope-Intercept Form

Graph the equation [latex] \displaystyle{y = -\frac{3}{4}x - 2} [/latex].

Solution

The equation is in the form [latex] y = mx + b [/latex].

Therefore, [latex] \displaystyle{m = -\frac{3}{4}} [/latex] and [latex] b = -2 [/latex], which means the y-intercept is the point (0, –2) and the slope is

[latex] \displaystyle{m = \frac{Rise}{Run} = \frac{Change in y value}{Change in x value} = \frac{-3}{4}} [/latex] or [latex] \displaystyle{\frac{3}{-4}} [/latex]

First plot the point (0, –2). Then, using the slope, [latex] \displaystyle{m = \frac{-3}{4}} [/latex], from the point (0, –2), move 4 units to the right and 3 units down to locate the new point, (4, –5).

Alternatively, using the slope, [latex] \displaystyle{m = \frac{3}{-4}} [/latex], from the point (0, –2), move 4 units to the left and 3 units up to locate another point on the line, (–4, 1).

Draw a line through these points to graph the equation.

Slope and Direction of a Line

The slope, m, of a line is a number that describes the direction of the line and the steepness of the slope.

The direction of the line either a. slopes upwards to the right, b. slopes downwards to the right, c. is horizontal, or d is vertical, as discussed below:

1. If the sign of 'm' is positive, then the line slopes upwards to the right (i.e., the line rises from left to right), as illustrated in Exhibit 8.2-e.

2. If the sign of 'm' is negative, then the line slopes downwards to the right (i.e., the line falls from left to right), as illustrated in Exhibit 8.2-f.

   
   
   

3. If 'm' is zero, then the line is horizontal (parallel to the X-axis). A slope of zero means that when the x-coordinate increases or decreases, the y-coordinate does not change (i.e., 'rise' = 0). This is a special case of the linear equation [latex] Ax + By = C [/latex], where the value [latex] A = 0 [/latex].

   Therefore, the equation of a horizontal line will be in the form y = b, and the y-intercept of the line is (0, b).

   For example, in the equation, [latex] y = 3 [/latex], (i.e., [latex] y = 0x + 3 [/latex]), the slope is zero and the value of the y-coordinate is 3 for all values of x.

   Therefore, the line is horizontal and passes through the y-intercept (0, 3), as illustrated in Exhibit 8.2-g.

   
   

4. If 'm' is undefined, then the line is vertical (parallel to the Y-axis). An undefined slope means that when the y-coordinate increases or decreases, the x-coordinate does not change (i.e., 'run' = 0). This is a special case of the linear equation [latex] Ax + By = C [/latex], where the value [latex] B = 0 [/latex].

   An equation in this form cannot be rearranged into slope-intercept form, as it would require dividing by 0. As such, rather than isolating for y, we isolate for the variable x.

   Therefore, the equation of a vertical line will be in the form [latex] x = a [/latex], and the x-intercept of the line is ([latex] a, 0 [/latex]).

   For example, in the equation, [latex] \boldsymbol{x = 2} [/latex], the slope is undefined and the value of the x-coordinate is 2 for all values of y.

   Therefore, the line is vertical and passes through the x-intercept (2, 0), as illustrated in Exhibit 8.2-h.

   
   

Slope and Steepness of a Line

The steepness of the line is also measured by the slope ('m') of the line. The farther the coefficient 'm' is away from zero, the steeper is the slope. The closer the coefficient 'm' is to zero, the flatter is the slope.

• A greater value for a positive slope indicates a steeper rise.

  
  

• A lesser value for a negative slope indicates a steeper fall.

  
  

• As discussed earlier, a slope of 0 indicates a horizontal line (no steepness) and an undefined slope indicates a vertical line (infinite steepness).

  
  

Lines Passing Through the Origin

Lines passing through the origin means that the point (0, 0) is on the line. That is, the coordinates of both the x- and y-intercepts are (0, 0). Hence, the equation of a line passing through the origin will be in the form [latex] y = mx [/latex], with the exception of two special cases:

• [latex] y = 0 [/latex] - the equation of the X-axis (i.e., the horizontal line passing through the origin)

• [latex] x = 0 [/latex] - the equation of the Y-axis (i.e., the vertical line passing through the origin)

For example, in the equation, [latex] \boldsymbol{y = 2x} [/latex], [latex] m = 2 [/latex], and [latex] b = 0 [/latex]; hence the slope is 2 and the y-intercept is the point (0, 0).

Similarly, in the equation [latex] \boldsymbol{y = -x} [/latex], [latex] m = -1 [/latex], and [latex] b = 0 [/latex]; hence the slope is –1 and the y-intercept is the point (0, 0).

Therefore, both of the lines above pass through the origin (0, 0), as illustrated in Exhibit 8.2-i.

Determining Equations of Lines

In order to determine the equation of a line, we need both the slope and the y-intercept. If we do not know this information, we can determine it by following these steps:

Step 1:

If the slope is unknown, determine it by computing the rise and run between two points on the graph.

Step 2:

Replace the unknown m in the slope-intercept form of the equation with the slope value from Step 1.

Step 3:

If the y-intercept is unknown, determine it by substituting the coordinates of a point on the line in for the x and y values in the slope-intercept form of the equation from Step 2, and solve for b.

Step 4:

Replace the unknown b in the slope-intercept form of the equation from Step 2 with the y-intercept value from Step 3 to arrive at the final equation of the line.

Example 8.2-e: Finding the Equation of a Line Given the Slope and One Point

Find the equation of a line having a slope of −2 and passing through the point (3, 5).

Solution

Step 1:

In this case, the slope is already known: [latex] m = -2 [/latex].

Step 2:

Replace m in the equation [latex] y = mx + b [/latex] with the value given.

Substituting for m in the slope-intercept equation [latex] y = mx + b [/latex], we obtain [latex] y = -2x + b [/latex].

Step 3:

Substitute the coordinates of the given point into the equation to solve for b.

Substituting the x- and y-coordinates of the point (3, 5) into the above equation, we obtain,

[latex] 5 = -2(3) + b [/latex]

[latex] b = 5 + 6 = 11 [/latex]

Step 4:

Write the equation in slope-intercept form, [latex] y = mx + b [/latex], by substituting for the values of m and b determined in the steps above.

Therefore, the equation of the line is [latex] y = -2x + 11 [/latex].

Example 8.2-f: Finding the Slope and the Equation of a Line Given Two Points

Find the equation of a line that passes through the points (3, 2) and (4, 5).

Solution

Step 1:

Calculate the slope.

[latex] \displaystyle{m = \frac{Change in y value}{Change in x value} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 2}{4 - 3} = \frac{3}{1} = 3} [/latex]

Step 2:

Replace m in the equation [latex] y = mx + b [/latex] with the calculated slope.

Substituting for m in the slope-intercept equation [latex] y = mx + b [/latex], we obtain [latex] y = 3x + b [/latex].

Step 3:

Substitute the coordinates of one of the given points into the equation to solve for b.

Substituting the x- and y-coordinates of the point (3, 2) into the above equation, we obtain,

[latex] 2 = 3(3) + b [/latex]

[latex] b = 2 - 9 = -7 [/latex]

Step 4:

Write the equation in slope-intercept form, [latex] y = mx + b [/latex], by substituting for the values of m and b determined in the steps above.

Therefore, the equation of the line is [latex] y = 3x - 7 [/latex].

Note: A good check to validate the equation is to substitute the coordinates of the other point into the equation to ensure that it is a solution.
That is, substituting the x- and y-coordinates of the point (4, 5) into the equation,

[latex] y = 3x - 7 [/latex]

[latex] 5 = 3(4) - 7 [/latex]

[latex] = 12 - 7 [/latex]

[latex] = 5 [/latex] (True)

Example 8.2-g: Finding the Equation of a Line Given a Graph

Determine the equation of the line in standard form that is plotted on the graph shown:

Solution

Step 1:

Start by choosing any two points (with integer coordinates) on the line: e.g., (1, 6) and (4, 10).

The slope of the line is,   [latex] \displaystyle{m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{10 - 6}{4 - 1} = \frac{4}{3}} [/latex].

Step 2:

Let the equation of the line be   [latex] y = mx + b [/latex]

Substituting   [latex] \displaystyle{m = \frac{4}{3}} [/latex],

Therefore,   [latex] \displaystyle{y = \frac{4}{3}x + b} [/latex]

Step 3:

Substituting the coordinates of one of the points (1, 6) into the above equation and solving for b,

[latex] \displaystyle{y = \frac{4}{3}x + b} [/latex]

[latex] \displaystyle{6 = \frac{4}{3}(1) + b} [/latex]

[latex] \displaystyle{b = 6 - \frac{4}{3} = \frac{18 - 4}{3} = \frac{14}{3}} [/latex]

Step 4:

Therefore, the equation of the line in slope-intercept form is: [latex] \displaystyle{y = \frac{4}{3}x + \frac{14}{3}} [/latex]

[latex] \displaystyle{y = \frac{4}{3}x + \frac{14}{3}} [/latex]   Multiplying each term by 3 and simplifying,

[latex] 3y = 4x + 14 [/latex]   Rearranging, ensuring that the coefficient of the x term remains positive,

[latex] 4x - 3y = -14 [/latex]

Therefore, the equation of the line in standard form is: [latex] 4x - 3y = -14 [/latex].

Parallel and Perpendicular Lines

Parallel Lines

Lines that never intersect are called parallel lines. All lines with the same slope are parallel to each other.

For example,

1. Lines represented by the equations [latex] \displaystyle{y = \frac{3}{2}x + 6} [/latex], [latex] \displaystyle{y = \frac{3}{2}x + 3} [/latex], and [latex] \displaystyle{y = \frac{3}{2}x - 6} [/latex] have the same slope (equal to [latex] \displaystyle{\frac{3}{2}} [/latex]).

   Therefore, they are all parallel to each other.

   
   

   All horizontal lines are parallel to each other and all vertical lines are parallel to each other. For example,

2. Lines represented by the equations [latex] y = 4 [/latex], [latex] y = 2 [/latex], and [latex] y = -2 [/latex] are horizontal lines and have slopes equalling zero. Therefore, they are all parallel to each other. Horizontal lines are also parallel to the X-axis (which has the equation [latex] y = 0 [/latex]).

   
   

3. Lines represented by the equations [latex] x = -4 [/latex], [latex] x = 2 [/latex], and [latex] x = 5 [/latex] are vertical lines and have undefined slopes. Therefore, they are all parallel to each other. Vertical lines are also parallel to the Y-axis (which has the equation [latex] x = 0 [/latex]).

   
   

Example 8.2-h: Writing the Equation of a Line Parallel to a Given Line and Passing Through a Given Point

Write the equation of the line that is parallel to [latex] 3x + y = 5 [/latex] and passes through the point P(2, 4).

Solution

[latex] 3x + y = 5 [/latex]   Rearranging to slope-intercept form,

[latex] y = -3x + 5 [/latex]   Therefore, the slope is [latex] m = -3 [/latex].

The slope of the line parallel to this will have the same slope, [latex] m = -3 [/latex].

Let the equation of the line parallel to [latex] 3x + y = 5 [/latex] be [latex] y= mx + b [/latex]. It passes through (2, 4) and has a slope of [latex] m = -3 [/latex].

[latex] y = mx + b [/latex]   Substituting the slope [latex] m = -3 [/latex],

[latex] y = -3x + b [/latex]   Substituting the coordinates of the given point (2, 4),

[latex] 4 = -3(2) + b [/latex]   Solving for [latex] b [/latex],

[latex] 4 = -6 + b [/latex]

[latex] b = 4 + 6 = 10 [/latex]

Therefore, the equation of the desired line is [latex] y = -3x + 10 [/latex], or [latex] 3x + y = 10 [/latex] in standard form.

Note: For parallel lines in standard form [latex] Ax + By = C [/latex], the values of [latex] A [/latex] and [latex] B [/latex] will always be proportional, and often will be equal.

Perpendicular Lines

Two lines that meet at a right angle are known as perpendicular lines. If the product of the slopes of two lines is −1, then the two lines are perpendicular to each other. This is the same as stating that if the slope of one line is the negative reciprocal of the other, then the two lines are perpendicular to each other.

For example,

1. Lines represented by the equations [latex] y = 2x + 4 [/latex] and [latex] \displaystyle{y = -\frac{1}{2}x + 1} [/latex] are perpendicular to each other because their slopes are negative reciprocals of each other.

   [latex]m_1 = 2 [/latex], [latex] \displaystyle{m_2 = -\frac{1}{2}} [/latex]

   [latex] \displaystyle{m_1 \cdot m_2 = 2\left(\frac{1}{2}\right) = -1} [/latex]

   
   

   Exhibit 8.2-n Vertical and Horizontal Lines are Perpendicular

   For example,

2. Lines represented by the equations [latex] y = 2x + 4 [/latex] and [latex] \displaystyle{y = -\frac{1}{2}x + 1} [/latex] are perpendicular to each other because their slopes are negative reciprocals of each other.

   
   

Example 8.2-i: Writing the Equation of a Line Perpendicular to a Given Line and Passing Through a Given Point

Write the equation of the line that is perpendicular to [latex] x + 3y = 9 [/latex] and passes through the point P(-4, 2).

Solution

[latex] x + 3y = 9 [/latex]   Rearranging to slope-intercept form,

[latex] 3y = -x + 9 [/latex]

[latex] \displaystyle{3y = -\frac{1}{3}x + 3} [/latex]   Therefore, the slope is [latex] \displaystyle{m = -\frac{1}{3}} [/latex].

The slope of the line perpendicular to this will be the negative reciprocal of [latex] \displaystyle{m = -\frac{1}{3}} [/latex], which is 3.

Let the equation of the line perpendicular to [latex] x + 3y = 9 [/latex] be [latex] y = mx + b [/latex]. It passes through the point (−4, 2) and has a slope of [latex] m = 3 [/latex].

[latex] y = mx + b [/latex]   Substituting the slope [latex] m = 3 [/latex],

[latex] y = 3x + b [/latex]   Substituting the coordinates of the given point (–4, 2),

[latex] 2 = 3(-4) + b [/latex]   Solving for [latex] b [/latex],

[latex] 2 = -12 + b [/latex]

[latex] 2 = -12 + b [/latex]

Therefore, the equation of the desired line is [latex] y = 3x + 14 [/latex], or [latex] 3x - y = -14 [/latex] in standard form.

8.2 Exercises

1. For the equation [latex] 2x + 3y = 18 [/latex], find the missing values in the following ordered pairs:

   a. (3, ?)   b. (-6, ?)   c. (0, ?)

   d. (?, 0)   e. ( ?, -4)   f. (?, 2)

   
   

2. For the equation [latex] x + 5y = 20 [/latex], find the missing values in the following ordered pairs:

   a. (0, ?)   b. (-15, ?)   c. (5, ?)

   d. (?, 6)   e. (?, -3)   f. (?, 0)

   
   

3. For the equation [latex] \displaystyle{y = -\frac{2}{3}x + 1} [/latex], find the missing values in the following ordered pairs:

   a. (6, ?) b. (–3, ?) c. (0, ?)

   d. (?, 0) e. (?, −3) f. (?, 5)

   
   

4. For the equation [latex] \displaystyle{y = -\frac{2}{3}x + 1} [/latex], find the missing values in the following ordered pairs:

   a. (2, ?)   b. (–3, ?)   c. (0, ?)

   d. (?, 0)   e. (?, −3)   f. (?, 9)

   
   

For Problems 5 to 10, write the equations in standard form.

5. [latex] \displaystyle{y = \frac{5}{2}x + 1} [/latex]

   
   

6. [latex] \displaystyle{y = \frac{2}{5}x - 1} [/latex]

   
   

7. [latex] \displaystyle{y = -\frac{3}{4}x - 3} [/latex]

   
   

8. [latex] \displaystyle{y = -\frac{4}{3}x + 4} [/latex]

   
   

9. [latex] \displaystyle{y = \frac{1}{2}x + \frac{3}{2}} [/latex]

   
   

10. [latex] \displaystyle{y = \frac{3}{2}x - \frac{1}{4}} [/latex]

    
    

For Problems 11 to 16, write the equations in slope-intercept form.

11. [latex] 4y + 6x = -3 [/latex]

    
    

12. [latex] 9y + 2x = 18 [/latex]

    
    

13. [latex] 3y - 2x = 15 [/latex]

    
    

14. [latex] 5y - 2x = -20 [/latex]

    
    

15. [latex] \displaystyle{\frac{x}{2} + \frac{y}{3} = 1} [/latex]

    
    

16. [latex] \displaystyle{\frac{x}{4} + \frac{y}{5} = 2} [/latex]

    
    

For Problems 17 to 24, graph the equations using a table of values.

17. [latex] y = x + 3 [/latex]

    
    

18. [latex] y = 3x + 2 [/latex]

    
    

19. [latex] y = -5x + 1 [/latex]

    
    

20. [latex] y = -2x + 3 [/latex]

    
    

21. [latex] 2x + y + 1 = 0 [/latex]

    
    

22. [latex] 4x + y + 2 = 0 [/latex]

    
    

23. [latex] 2x - y - 3 = 0 [/latex]

    
    

24. [latex] x - y - 1 = 0 [/latex]

    
    

For Problems 25 to 30, determine the x- and y-intercepts for the equations and graph the equations.

25. [latex] 3x + y = -2 [/latex]

    
    

26. [latex] 5x + y = -3 [/latex]

    
    

27. [latex] x + y - 3 = 4 [/latex]

    
    

28. [latex] x + y - 4 = 7 [/latex]

    
    

29. [latex]y = 4x + 1 [/latex]

    
    

30. [latex] 4x + y + 2 = 0 [/latex]

    
    

For Problems 31 to 34, determine the slopes and y-intercepts of the equations and graph the equations.

31. [latex] 2x - 3y - 18 = 0 [/latex]

    
    

32. [latex] 5x - 2y + 10 = 0 [/latex]

    
    

33. [latex] -4x + 7y - 21 = 0 [/latex]

    
    

34. [latex] -7x + 8y - 32 = 0 [/latex]

    
    

35. Point 'A' is in the 3^rd quadrant and Point 'B' is in the 1^st quadrant. Determine the sign of the slope of the line AB.

    
    

36. Point 'C' is in the 4^th quadrant and Point 'D' is in the 2^nd quadrant. Determine the sign of the slope of the line CD.

    
    

For Problems 37 to 40, determine the slopes of the lines passing through:

37. (2, 1) and (6, 1)

    
    

38. (−6, 4) and (2, 4)

    
    

39. (−5, 4) and (3, −1)

    
    

40. (5, 6) and (5, −4)

    
    

For Problems 41 to 44, determine the equations of the lines in slope-intercept form that pass through the points.

41. (1, 2) and (5, 2)

    
    

42. (5, 0) and (4, 5)

    
    

43. (−3, −5) and (3, 1)

    
    

44. (−4, −7) and (5, 2)

    
    

For Problems 45 to 50, determine the equations of the lines in slope-intercept form that:

45. Have a slope of −2 and pass through (2, 6).

    
    

46. Have a slope of 3 and pass through (−3, −2).

    
    

47. Have a slope of [latex] \displaystyle{\frac{2}{3}} [/latex] and pass through the origin.

    
    

48. Have a slope of [latex] \displaystyle{-\frac{4}{5}} [/latex] and pass through the origin.

    
    

49. Have an x-intercept = 4 and a y-intercept = 6.

    
    

50. Have an x-intercept = −4 and a y-intercept = 2.

    
    

51. The slope of a line is 3. The line passes through A(4, y) and B(6, 8). Find y.

    
    

52. The slope of a line is 2. The line passes through A(x, 8) and B(2, 4). Find x.

    
    

53. Points A(2, 3), B (6, 5), and C(10, y) are on a line. Find y.

    
    

54. Points D(3, 2), E (6, 5), and F(x, 1) are on a line. Find x.

    
    

For Problems 55 to 58, determine the equation of the line (in standard form) for the graphs.

For Problems 59 to 66, use the slope property of parallel and perpendicular lines to identify whether each of the pairs of lines are parallel, perpendicular, or neither.

59. [latex] y = x + 1 [/latex]

    [latex] 4x + 4y = -1 [/latex]

    
    

60. [latex] 6x - 5y = 10 [/latex]

    [latex] \displaystyle{y = -\frac{6}{5}x - 12} [/latex]

    
    

61. [latex] x - 3y = -60 [/latex]

    [latex] \displaystyle{y = \frac{1}{3}x - 4} [/latex]

    
    

62. [latex] 3x - 2y = -12 [/latex]

    [latex] 2x + 3y = -12 [/latex]

    
    

63. [latex] 2x + 5y = -5 [/latex]

    [latex] \displaystyle{y = \frac{5}{2}x - 4} [/latex]

    
    

64. [latex] 7x + 4y = 16 [/latex]

    [latex] \displaystyle{y = -\frac{4}{7}x + 3} [/latex]

    
    

65. [latex] 3x - 2y = -6 [/latex]

    [latex] \displaystyle{y = \frac{3}{2}x - 16} [/latex]

    
    

For Problems 67 to 74, determine the equations for the lines in slope-intercept form.

67. A line parallel to [latex] 3y - 2x = 6 [/latex] and passing through the point P(2, −3).

    
    

68. A line parallel to [latex] y = 3x - 1 [/latex] and passing through the point P(−2, −4).

    
    

69. A line parallel to [latex] 3x - 9y = -2 [/latex] and passing through the y-intercept of the line [latex] 5x - y = 20 [/latex].

    
    

70. A line parallel to [latex] 3x + y = -2 [/latex] and passing through the x-intercept of the line [latex] 2x + 3y - 4 = 0 [/latex].

    
    

71. A line perpendicular to [latex] x + y = 3 [/latex] and passing through the point P(−2, 5).

    
    

72. A line perpendicular to [latex] 4x + y + 1 = 0 [/latex] and passing through the point P(3, 4).

    
    

73. A line perpendicular to [latex] 2x - y = 5 [/latex] and passing through the x-intercept of the line [latex] 3x + 2y - 6 = 0 [/latex].

    
    

74. A line perpendicular to [latex] 3x + y + 9 = 0 [/latex] and passing through the y-intercept of the line [latex] 2x + 3y - 10 = 0 [/latex].