Slope-Intercept Form (2024)

The slope-intercept is the most “popular” form of a straight line. Many students find this useful because of its simplicity. One can easily describe the characteristics of the straight line even without seeing its graph because the slope and [latex]y[/latex]-intercept can easily be identified or read off from this form.

Slope-Intercept Form of the Equation of a Line

The linear equation written in the form

[latex]\large{y = mx + b}[/latex]

is in slope-intercept formwhere:

[latex]m[/latex] is the slope, and [latex]b[/latex] is the [latex]y[/latex]-intercept

Slope-Intercept Form (1)

Quick notes:

  • The slope [latex]m[/latex] measures how steep the line is with respect to the horizontal. Given two points [latex]\left( {{x_1},{y_1}} \right)[/latex] and [latex]\left( {{x_2},{y_2}} \right)[/latex] found in the line, the slope is computed as
Slope-Intercept Form (2)
  • The [latex]y[/latex]-intercept [latex]b[/latex] is the point where the line crosses the [latex]y[/latex]-axis. Notice that in the graph below, the red dot is always found on the main vertical axis of the Cartesian plane. That is the basic characteristic of the [latex]y[/latex]-intercept.
Slope-Intercept Form (3)

Let’s go over some examples of how to write the equation of a straight line in linear form [latex]y = mx + b[/latex].

Examples of Applying the Concept of Slope-Intercept Form of a Line

Example 1: Write the equation of the line in slope-intercept form with a slope of[latex] – \,5[/latex] and a [latex]y[/latex]-intercept of [latex]3[/latex].

The needed information to write the equation of the line in the form [latex]y = mx + b[/latex] are clearly given in the problem since

[latex]m = – \,5[/latex] (slope)

[latex]b = 3[/latex] ([latex]y[/latex]-intercept)

Substituting in [latex]y = mx + b[/latex], we obtain

Slope-Intercept Form (4)

By having a negative slope, the line is decreasing/falling fromleft to right, and passing through the [latex]y[/latex]-axis at point [latex]\left( {0,3} \right)[/latex].

Slope-Intercept Form (5)

Example 2: Write the equation of the line in slope-intercept form with a slope of [latex]7[/latex] and a [latex]y[/latex]-intercept of[latex] – \,4[/latex].

The slope is given as [latex]m = 7[/latex] and the [latex]y[/latex]-intercept as [latex]b = – \,4[/latex]. Substituting into the slope-intercept formula [latex]y = mx + b[/latex], we have

Slope-Intercept Form (6)

The slope is positivethus the line is increasing or rising from left to right, but passing through the [latex]y[/latex]-axis at point [latex]\left( {0, – \,4} \right)[/latex].

Slope-Intercept Form (7)

Example 3: Write the equation of the line in slope-intercept with a slope of [latex]9[/latex] and passing through the point [latex]\left( {0, – \,2} \right)[/latex].

This problem is slightly different from the previous two examplesbecause the [latex]y[/latex]-intercept [latex]b[/latex] is not given to us upfront. So our next goal is to somehow figure out the value of [latex]b[/latex] first.

However, if we examine the slope-intercept form, it shouldlead us to believe that we have enough information to solve for [latex]b[/latex].How?

Slope-Intercept Form (8)

That means [latex]m = 9[/latex], and from the given point [latex]\left( {0, – \,2} \right)[/latex] we have [latex]x = 0[/latex] and [latex]y = – \,2[/latex].Let’s substitute these known values into the slope-intercept formula and solve for the missing value of [latex]b[/latex].

Slope-Intercept Form (9)

Now it is possible to write the slope-intercept form as

Slope-Intercept Form (10)

Example 4: Find the slope-intercept form of the line with a slope of[latex] – \,3[/latex] and passing through the point [latex]\left( { – 1,\,15} \right)[/latex].

Again, the value of [latex]y[/latex]-intercept [latex]b[/latex] is not directly provided to us. But we can utilize the given slope and a point to find it.

Slope-Intercept Form (11)

Substitute the known values into the slope-intercept formula, and then solve for the unknown value of [latex]b[/latex].

Slope-Intercept Form (12)

Back substitute the value of the slope and the solved value of the [latex]y[/latex]-intercept into [latex]y = mx + b[/latex].

Slope-Intercept Form (13)

Example 5: A line with the slope of[latex] – \,8[/latex] and passing through the point [latex]\left( { – \,4,\, – 1} \right)[/latex].

The given slope is [latex]m = – \,8[/latex] and from the given point [latex]\left( { – \,4,\, – 1} \right)[/latex], we have [latex]x = – \,4[/latex] and [latex]y = – \,1[/latex].Now, we are going to substitute the known values into the slope-intercept form of the lineto solve for [latex]b[/latex].

Slope-Intercept Form (14)

Since [latex]m = – \,8[/latex] and [latex]b = – \,33[/latex],the slope-intercept form of the line becomes

Slope-Intercept Form (15)

Example 6: Write the slope-intercept form of the line with a slope of[latex]{3 \over 5}[/latex] and through the point [latex]\left( {5,\, – 2} \right)[/latex].

We have a slope here that is notan integer, i.e. the denominator is other than positive or negative one,[latex] \pm 1[/latex]. In other words, we have a“true” fractional slope.

The procedure for solving this problem is very similar to examples #3, #4, and #5.But the main point of this example is to emphasize the algebraicsteps required on how to solve a linear equation involving fractions.

The known values ofthe problemare

  • Given slope:
Slope-Intercept Form (16)
  • Given point:
Slope-Intercept Form (17)

Plug thevalues into [latex]y = mx + b[/latex] and solve for [latex]b[/latex].

Slope-Intercept Form (18)

As you can see the common factors of [latex]5[/latex] in the numerator and denominator nicely cancel each other out which greatly simplifies the process of solving for [latex]b[/latex].

Putting this togetherin the form [latex]y = mx + b[/latex]

Slope-Intercept Form (19)

Example 7: Slope of[latex]{{\, – 3} \over 2}[/latex] and through the point [latex]\left( { – 1,\, – 1} \right)[/latex].

The given slope is [latex]m = {{\, – 3} \over 2}[/latex] and from the given point[latex]\left( { – 1,\, – 1} \right)[/latex],the values of [latex]x[/latex] and [latex]y[/latex] can easily be identified.

Slope-Intercept Form (20)

Now plug in the known values into the slope-intercept form [latex]y = mx + b[/latex] to solve for [latex]b[/latex].

Slope-Intercept Form (21)

Make sure that when you add or subtract fractions, you generate a common denominator.

Slope-Intercept Form (22)

After getting the value of [latex]b[/latex],we can now write the slope-intercept form of the line.

Slope-Intercept Form (23)

Example 8: Slope of[latex] – \,6[/latex] and through the point [latex]\left( {{1 \over 2},{1 \over 3}} \right)[/latex].

The slope is given as [latex]m = – \,6[/latex] and from the point, we have [latex]x = {1 \over 2}[/latex] and [latex]y = {1 \over 3}[/latex].

Substitute the known values into [latex]y = mx + b[/latex]. Then solve the missing value of [latex]b[/latex].

Slope-Intercept Form (24)

Therefore, the slope-intercept form of the line is

Slope-Intercept Form (25)

Example 9: Slope of[latex]{{\,7} \over 3}[/latex] andthrough the point[latex]\left( {{{ – \,2} \over 5},{5 \over 2}} \right)[/latex].

Identifying the known values

  • Given slope:
Slope-Intercept Form (26)
  • Given point:
Slope-Intercept Form (27)

The setup to find [latex]b[/latex] becomes

Slope-Intercept Form (28)

That makes the slope-intercept form of the line as

Slope-Intercept Form (29)

Example 10: A line passing through the given two points [latex]\left( {4,\,5} \right)[/latex] and [latex]\left( {0,\,3} \right)[/latex].

In this problem, we are not provided with both the slope [latex]m[/latex] and [latex]y[/latex]-intercept [latex]b[/latex].However, weshould realize that the slope is easily calculated when two points are known using the Slope Formula.

Slope Formula

The slope, [latex]m[/latex], of a line passing through two arbitrary points [latex]\left( {{x_1},{y_1}} \right)[/latex] and [latex]\left( {{x_2},{y_2}} \right)[/latex] is calculated as follows…

Slope-Intercept Form (30)

If we let [latex]\left( {4,\,5} \right)[/latex] be the first point, then [latex]\left( {0,\,3} \right)[/latex] must be the second.

Labeling the components of each point should help in identifying the correct values that would be substituted into the slope formula.

Slope-Intercept Form (31)

Based on the labeling above, now we know that

Slope-Intercept Form (32)

Next, write the slope formula, plug in the known valuesand simplify.

Slope-Intercept Form (33)

Great! We found the slope to be [latex]m = {{\,1} \over 2}\,[/latex].The only missing piece of the puzzle is to determine the [latex]y[/latex]-intercept. Use theslope that we found, together with ANY of the two given points. In this exercise, I will show you that we should arrive at the same value of the [latex]y[/latex]-intercept regardless of which point is selected for the calculation.

Finding the [latex]y[/latex]-intercept

  • Using the first point [latex]\left( {4,\,5} \right)[/latex].
Slope-Intercept Form (34)
Slope-Intercept Form (35)
  • Using the second point [latex]\left( {0,\,3} \right)[/latex].
Slope-Intercept Form (36)
Slope-Intercept Form (37)

Indeed, the [latex]y[/latex]-intercepts come out the same in both calculations.We can now write the linear equation in slope-intercept form.

Slope-Intercept Form (38)

Below is the graph of the line passing through the given two points.

Slope-Intercept Form (39)

Example 11: A line passing through the given two points [latex]\left( { – \,7,\,4} \right)[/latex] and [latex]\left( { – \,2,\,19} \right)[/latex].

Let’s solve this step by step.

  • Step 1: Assign which point is the first and second, and then label its components.
Slope-Intercept Form (40)
  • Step 2: Substitute the known values into the slope formula, and simplifyif necessary.
Slope-Intercept Form (41)
  • Step 3: Pick any of the two given points. Suppose we pick the point [latex]\left( { – \,7,\,4} \right)[/latex].That means [latex]x = – \,7[/latex] and [latex]y = 4[/latex].Using the calculated value of slope in step 2, we can now find the [latex]y[/latex]-intercept [latex]b[/latex].
Slope-Intercept Form (42)
  • Step 4: Putting them together in [latex]y = mx + b[/latex] form, since [latex]m = 3[/latex] and [latex]b = 25[/latex],we have the slope-intercept form of the line as
Slope-Intercept Form (43)
  • Step 5:Using a graphing utility, show that the solved linear equation in slope-intercept form passes through the two points.
Slope-Intercept Form (44)

Example 12: A line passing through the given two points [latex]\left( { – \,6,\, – \,3} \right)[/latex] and [latex]\left( { – \,7,\, – 1} \right)[/latex].

  • Find the slope
Slope-Intercept Form (45)
Slope-Intercept Form (46)
  • Pick any of the two given points. Suppose, we chose the second point which is
Slope-Intercept Form (47)

Substitute known values in the slope-intercept form [latex]y = mx + b[/latex] to solve for [latex]b[/latex].

Slope-Intercept Form (48)
  • Putting them together. Since [latex]m = – \,2[/latex] and [latex]b = – \,15[/latex],the slope-intercept form of the line is
Slope-Intercept Form (49)
  • This is the graph of the line showing that it passes both ofthe two points.
Slope-Intercept Form (50)

Example 13: A line passing through the given two points [latex]\left( {5,\, – \,2} \right)[/latex] and [latex]\left( { – \,2,\,5} \right)[/latex].

  • Determine the slope from the given two points
Slope-Intercept Form (51)
Slope-Intercept Form (52)
  • Pick any of the two given points. Let’s say we chose the first one, [latex]\left( {5,\, – \,2} \right)[/latex]. That means [latex]x = 5[/latex], and [latex]y = – \,2[/latex]. Use this information together with the value of slope to solve for the [latex]y[/latex]-intercept [latex]b[/latex].
Slope-Intercept Form (53)
  • Now, put them together. Since [latex]m = – \,1[/latex] and [latex]b = 3[/latex], the slope-intercept form of the line is
Slope-Intercept Form (54)
  • Using a graphing utility, show that the line passes through the two given points.
Slope-Intercept Form (55)

You might also like these tutorials:

  • Types of Slopes of a Line
  • Slope Formula of a Line
  • Point-Slope Form of a Line

Tags: Intermediate Algebra, Lessons

Slope-Intercept Form (2024)

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