Choose the Most Convenient Method to Graph a Line
While we could graph any linear equation by plotting points, it may not always be the most convenient method. This table shows six of equations we’ve graphed in this chapter, and the methods we used to graph them.

Equation 
Method 
#1 
y = 2x + 1 
Plotting points 
#2 
y = \(\dfrac{1}{2}\)x + 3 
Plotting points 
#3 
x = −7 
Vertical line 
#4 
y = 4 
Horizontal line 
#5 
2x + y = 6 
Intercepts 
#6 
4x − 3y = 12 
Intercepts 
What is it about the form of equation that can help us choose the most convenient method to graph its line?
Notice that in equations #1 and #2, y is isolated on one side of the equation, and its coefficient is 1. We found points by substituting values for x on the right side of the equation and then simplifying to get the corresponding y values.
Equations #3 and #4 each have just one variable. Remember, in this kind of equation the value of that one variable is constant; it does not depend on the value of the other variable. Equations of this form have graphs that are vertical or horizontal lines.
In equations #5 and #6, both x and y are on the same side of the equation. These two equations are of the form Ax + By = C . We substituted y = 0 and x = 0 to find the x and y intercepts, and then found a third point by choosing a value for x or y.
This leads to the following strategy for choosing the most convenient method to graph a line.
HOW TO: CHOOSE THE MOST CONVENIENT METHOD TO GRAPH A LINE
Step 1. If the equation has only one variable. It is a vertical or horizontal line.
 x = a is a vertical line passing through the xaxis at a.
 y = b is a horizontal line passing through the yaxis at b.
Step 2. If y is isolated on one side of the equation. Graph by plotting points.
 Choose any three values for x and then solve for the corresponding y values.
Step 3. If the equation is of the form Ax + By = C, find the intercepts.
 Find the x and y intercepts and then a third point.
Example \(\PageIndex{7}\):
Identify the most convenient method to graph each line: (a) y = −3 (b) 4x−6y = 12 (c) x = 2 (d) y = 2 5 x−1
Solution
(a) y = −3
This equation has only one variable, y. Its graph is a horizontal line crossing the yaxis at −3.
(b) 4x−6y = 12
This equation is of the form Ax + By = C. Find the intercepts and one more point.
(c) x = 2
There is only one variable, x. The graph is a vertical line crossing the xaxis at 2.
(d) y = \(\dfrac{2}{5}\)x−1
Since y is isolated on the left side of the equation, it will be easiest to graph this line by plotting three points.
Exercise \(\PageIndex{7A}\):
Identify the most convenient method to graph each line: (a) 3x + 2y = 12 (b) y = 4 (c) y = \(\dfrac{1}{5}\)x−4 (d) x = −7
 Answer a

intercepts
 Answer b

horizontal line
 Answer c

plotting points
 Answer d

vertical line
Exercise \(\PageIndex{7B}\):
Identify the most convenient method to graph each line: (a) x = 6 (b) y = \(− \dfrac{3}{4}\)x + 1 (c) y = −8 (d) 4x−3y = −1
 Answer a

vertical line
 Answer b

plotting points
 Answer c

horizontal line
 Answer d

intercepts
Practice Makes Perfect
Find the x and y Intercepts from an Equation of a Line
In the following exercises, find the intercepts.
 x + y = 4
 x + y = 3
 x + y = −2
 x + y = −5
 x − y = 5
 x − y = 1
 x − y = −3
 x − y = −4
 x + 2y = 8
 x + 2y = 10
 3x + y = 6
 3x + y = 9
 x−3y = 12
 x−2y = 8
 4x − y = 8
 5x − y = 5
 2x + 5y = 10
 2x + 3y = 6
 3x−2y = 12
 3x−5y = 30
 y = \(\dfrac{1}{3}\)x−1
 y = \(\dfrac{1}{4}\)x−1
 y = \(\dfrac{1}{5}\)x + 2
 y = \(\dfrac{1}{3}\) x + 4
 y = 3x
 y = −2x
 y = −4x
 y = 5x
Graph a Line Using the Intercepts
In the following exercises, graph using the intercepts.
 −x + 5y = 10
 −x + 4y = 8
 x + 2y = 4
 x + 2y = 6
 x + y = 2
 x + y = 5
 x + y = −3
 x + y = −1
 x − y = 1
 x − y = 2
 x − y = −4
 x − y = −3
 4x + y = 4
 3x + y = 3
 3x − y = −6
 2x − y = −8
 2x + 4y = 12
 3x + 2y = 12
 3x−2y = 6
 5x−2y = 10
 2x−5y = −20
 3x−4y = −12
 y = −2x
 y = −4x
 y = x
 y = 3x
Choose the Most Convenient Method to Graph a Line
In the following exercises, identify the most convenient method to graph each line.
 x = 2
 y = 4
 y = 5
 x = −3
 y = −3x + 4
 y = −5x + 2
 x − y = 5
 x − y = 1
 y = \(\dfrac{2}{3}\)x−1
 y = \(\dfrac{4}{5}\)x−3
 y = −3
 y = −1
 3x−2y = −12
 2x−5y = −10
 y = \(− \dfrac{1}{4}\)x + 3
 y = \(− \dfrac{1}{3}\)x + 5
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) What does this checklist tell you about your mastery of this section? What steps will you take to improve?