Graphing Linear Equations

Graphing linear equations allows us to represent relationships between variables on a coordinate plane visually. It helps us predict values, understand trends, and analyze mathematical relationships.

A linear equation with 2 variables always forms a straight line and is commonly written in slope-intercept form:

y = mx + c, where 

• m is the slope, representing the rate of change or steepness of the line.
• c  is the y-intercept, the point where the line crosses the y-axis.

Using X-Intercept and Y-Intercept Only

Let us graph the linear equation: y = x + 1

To find its intercepts:

Substituting x = -1, 

y = -1 + 1 = 0

So, the x-intercept is (-1, 0)

Now, setting x = 0, we get

y = 0 + 1 = 1

So, the y-intercept is (0,1)

By plotting these points on a graph, we will get the graph of the given linear equation.

However, if a linear equation is given in a form other than the slope-intercept form, it can be rearranged into the slope-intercept form y = mx + c and then graphed for simplicity.

Using Slope and Y-Intercept 

Another useful way to graph a linear equation is by using the slope and the y-intercept.

Let us graph the linear equation y = 2x + 1

Step 1: Identifying the Slope and Y-Intercept

Here, the slope (m) is 2, which means for every 1-unit increase in x (moving right along the x-axis), the value of y increases by 2 units (moving up on the y-axis).

The y-intercept (c) is 1, which is the point where the line crosses the y-axis.

Thus, the coordinates of the y-intercept is (0, 1)

Step 2: Plotting the Y-Intercept 

Now, we will plot the y-intercept as shown:

Step 3: Finding Some Others Coordinates and Making the Table

Now, by using the slope m = 2, we can find another point that will satisfy the equation.

The slope m = 2 can be written as a fraction: 

m = ${\dfrac{2}{1}}$

Here,​

2 is the rise (i.e., vertical movement), and 

1 is the run (i.e., horizontal movement).

From the point (0, 1), if we move 2 units up and 1 unit to the right, we will get the point (1, 3).

By putting the values of x and y in the table, we get:

x	2x + 1	y
-2	2(-2) + 1	-3
-1	2(-1) + 1	-1
0	2(0) + 1	1
1	2(1) + 1	3
2	2(2) + 1	5

Now, by plotting these points on the graph, 

Step 4: Joining the Points 

Now, by connecting all points on the graph, we get

Verifying The Coordinates as Solutions

Let us now verify whether the points we obtained above are the solutions to the given linear equation.

This will be done by substituting the values of x in the equation.

At x = -2, y = 2(-2) + 1 = -4 + 1 = -3

Thus, (-2, -3) is a solution of y = 2x + 1

At x = -1, y = 2(-1) + 1 = -2 + 1 = -1

Thus, (-1, -1) is a solution of y = 2x + 1

At x = 0, y = 2(0) + 1 = 1

Thus, (0, 1) is a solution of y = 2x + 1

At x = 1, y = 2(1) + 1 = 2 + 1 = 3

Thus, (1, 3) is a solution of y = 2x + 1

At x = 2, y = 2(2) + 1 = 4 + 1 = 5

Thus, (2, 5) is a solution of y = 2x + 1

Special Cases

Graphing Horizontal Line

A horizontal line is written as y = c, where c is a constant. 

For example, y = 1, which means y is always 1, regardless of x.

Graphing Vertical Line

A vertical line is written as x = c, where c is a constant. 

For example, x = -1, which means x is always -1, regardless of y.

Solved Examples

Example 1. Graph: 4x – y = 8

Given, 4x – y = 8 …..(i)

Converting the equation (i) to slope-intercept form,

-y = -4x + 8 

⇒ y = 4x – 8

Here, 

Slope = m = 4

Y-intercept = c = -8

Thus, the coordinates of the y-intercept are (0, -8)

Now, using the slope and the y-intercept, the other points are:

x	4x – 8	y
-2	4(-2) – 8	-16
-1	4(-1) – 8	-12
0	4(0) – 8	-8
1	4(1) – 8	-4
2	4(2) – 8	0

Plotting these points, we get the required graph.

Example 2. Graph: 5x + 2y = 10

Given, 5x + 2y = 10

Converting the equation (i) to slope-intercept form,

2y = -5x + 10

⇒ y = ${-\dfrac{5}{2}x+\dfrac{10}{2}}$

⇒ y = ${-\dfrac{5}{2}x+5}$

Here,

Slope = m = ${-\dfrac{5}{2}}$

Y-intercept = c = 5

Thus, the coordinates of the y-intercept are (0, 5)

Now, using the slope and the y-intercept, the other points are:

x	${-\dfrac{5}{2}x+5}$	y
-2	${\left( -\dfrac{5}{2}\right) \left( -2\right) +5}$	10
-1	${\left( -\dfrac{5}{2}\right) \left( -1\right) +5}$	7.5
0	${\left( -\dfrac{5}{2}\right) \left( 0\right) +5}$	5
1	${\left( -\dfrac{5}{2}\right) \left( 1\right) +5}$	2.5
2	${\left( -\dfrac{5}{2}\right) \left( 2\right) +5}$	0

When plotting these points, we get the required graph.

Example 3. Graph: y = -3x + 6

Given, y = -3x + 6

Here, 

Slope = m = -3

Y-intercept = c = 6

Thus, the coordinates of the y-intercept are (0, 6)

Now, using the slope and the y-intercept, the other points are:

x	-3x + 6	y
-2	-3(-2) + 6	12
-1	-3(-1) + 6	9
0	-3(0) + 6	6
1	-3(1) + 6	3
2	-3(2) + 6	0

When plotting these points, we get the required graph.

Example 4. Graph: ${y=-\dfrac{1}{2}x+3}$

Given, y = ${y=-\dfrac{1}{2}x+3}$

Here, 

Slope = m = ${-\dfrac{1}{2}}$

Y-intercept = c = 3

Thus, the coordinates of the y-intercept are (0, 3)

Now, using the slope and the y-intercept, the other points are:

x	${-\dfrac{1}{2}x+3}$	y
-4	${\left( -\dfrac{1}{2}\right) \left( -4\right) +3}$	5
-2	${\left( -\dfrac{1}{2}\right) \left( -2\right) +3}$	4
0	${\left( -\dfrac{1}{2}\right) \left( 0\right) +3}$	3
2	${\left( -\dfrac{1}{2}\right) \left( 2\right) +3}$	2
4	${\left( -\dfrac{1}{2}\right) \left( 4\right) +3}$	1

When plotting these points, we get the required graph.

Example 5. What is the equation of the line graphed below? Also, verify whether the point (1, 4) is a solution to the equation.

As we know, the equation of a line in slope-intercept form is y = mx + c

Here, m is the slope of the line, and c is the y-intercept.

Since the slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is:

${m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}}$

Given the points are (0, 5) and (-5, 0)

Now, 

m = ${\dfrac{0-5}{-5-0}}$ = 1

The y-intercept = c = 5

Thus, the equation of the line from the given graph is y = x + 5

Substituting x = 1 into the equation y = x + 5, we get

y = 1 + 5 = 6, which is not equal to 4

Thus, (1, 4) is not a solution to the equation y = x + 5.