When graphing a line we found one method we could use is to make a table of values. However, if we can identify some properties of the line, we may be able to make a graph much quicker and easier. One such method is finding the slope and the y-intercept of the equation. The slope can be represented by m and the y-intercept, where it crosses the axis and [latex]x=0[/latex], can be represented by [latex](0,b)[/latex] where b is the value where the graph crosses the vertical y-axis. Any other point on the line can be represented by [latex](x,y)[/latex].

In the equation,

[latex]y = mx + b[/latex]

[latex]\begin{array}{l}\,\,\,\,\,m\,\,\,\,=\,\,\,\text{slope}\\(x,y)=\,\,\,\text{a point on the line}\\\,\,\,\,\,\,\,b\,\,\,\,=\,\,\,\text{the y value of the y-intercept}\end{array}[/latex]

This formula is known as the slope-intercept equation. If we know the slope and the y-intercept we can easily find the equation that represents the line

We can also find the equation by looking at a graph and finding the slope and y-intercept.

We can also move the opposite direction, using the equation identify the slope and y-intercept and graph the equation from this information. However, it will be important for the equation to first be in slope intercept form. If it is not, we will have to solve it for y so we can identify the slope and the y-intercept.

Once we have an equation in slope-intercept form we can graph it by first plotting the y-intercept, then using the slope, find a second point and connecting the dots.

Slope-Intercept Form of a Line

Find the Equation of a Line Given the Slope and a Point on the Line

Using the slope-intercept equation of a line is possible when you know both the slope (m) and the y-intercept (b), but what if you know the slope and just any point on the line, not specifically the y-intercept? Can you still write the equation? The answer is yes, but you will need to put in a little more thought and work than you did previously.

Recall that a point is an (x, y) coordinate pair and that all points on the line will satisfy the linear equation. So, if you have a point on the line, it must be a solution to the equation. Although you don’t know the exact equation yet, you know that you can express the line in slope-intercept form, [latex]y=mx+b[/latex].

You do know the slope (m), but you just don’t know the value of the y-intercept (b). Since point (x, y) is a solution to the equation, you can substitute its coordinates for x and y in [latex]y=mx+b[/latex] and solve to find b!

This may seem a bit confusing with all the variables, but an example with an actual slope and a point will help to clarify.

To confirm our algebra, you can check by graphing the equation [latex]y=3x+1[/latex]. The equation checks because when graphed it passes through the point [latex](1,4)[/latex].

If you know the slope of a line and a point on the line, you can draw a graph. Using an equation in the point-slope form allows you to identify the slope and a point. Consider the equation [latex]\displaystyle y=-3x-1[/latex]. The y-intercept is the point on the line where it passes through the y-axis. What is the value of x at this point?

Therefore, you can tell from this equation that the y-intercept is at [latex](0,−1)[/latex], check this by replacing x with 0 and solving for y. To graph the line, start by plotting that point, [latex](0,−1)[/latex], on a graph.

You can also tell from the equation that the slope of this line is [latex]−3[/latex]. So start at [latex](0,−1)[/latex] and count up 3 and over [latex]−1[/latex] (1 unit in the negative direction, left) and plot a second point. (You could also have gone down 3 and over 1.) Then draw a line through both points, and there it is, the graph of [latex]\displaystyle y=-3x-1[/latex].

Video: Find the Equation of a Line Given the Slope and a Point on the Line

Find the Equation of a Line Given Two Points on the Line

Let’s suppose you don’t know either the slope or the y-intercept, but you do know the location of two points on the line. It is more challenging, but you can find the equation of the line that would pass through those two points. You will again use slope-intercept form to help you.

The slope of a linear equation is always the same, no matter which two points you use to find the slope. Since you have two points, you can use those points to find the slope (m). Now you have the slope and a point on the line! You can now substitute values for m, x, and y into the equation [latex]y=mx+b[/latex] and find b.

Notice that is doesn’t matter which point you use when you substitute and solve for b—you get the same result for b either way. In the example above, you substituted the coordinates of the point (2, 1) in the equation [latex]y=2x+b[/latex]. Let’s start with the same equation, [latex]y=2x+b[/latex], but substitute in [latex](−1,−5)[/latex]:

[latex]\begin{array}{l}\,\,\,\,\,y=2x+b\\-5=2\left(-1\right)+b\\-5=-2+b\\-3=b\end{array}[/latex]

The final equation is the same: [latex]y=2x–3[/latex].

Video: Find the Equation of a Line Given Two Points on the Line

Write the equations of parallel and perpendicular lines

The relationships between slopes of parallel and perpendicular lines can be used to write equations of parallel and perpendicular lines.

Let’s start with an example involving parallel lines.

Determine the Equation of a Line Parallel to Another Line Through a Given Point

Determine the Equation of a Line Perpendicular to Another Line Through a Given Point

When you are working with perpendicular lines, you will usually be given one of the lines and an additional point. Remember that two non-vertical lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. To find the slope of a perpendicular line, find the reciprocal, and then find the opposite of this reciprocal.  In other words, flip it and change the sign.

Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form

Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line

 Interpret the y-intercept of a linear equation

Often, when the line in question represents a set of data or observations, the y-intercept can be interpreted as a starting point.  We will continue to use the examples for house value in Mississippi and Hawaii and high school smokers to interpret the meaning of the y-intercept in those equations.

The examples that follow show how to interpret the y-intercept of the equations used to model house value and the number of high school smokers. Additionally, you will see how to use the equations to make predictions about house value and the number of smokers in future years.

In the following video you will see an example of how to interpret the y- intercept given a linear equation that represents a set of data.

Use a linear equation to make a prediction

Another useful outcome we gain from writing equations from data is the ability to make predictions about what may happen in the future. We will continue our analysis of the house price and high school smokers. In the following examples you will be shown how to predict future outcomes based on the linear equations that model current behavior.

In the following video, you will see the example of how to make a prediction with the home value data.

The following video gives a thorough explanation of making a prediction given a linear equation.

Bringing it Together

The last example we will show will include many of the concepts that we have been building up throughout this section.  We will interpret a word problem, write a linear equation from it, graph the equation, interpret the y-intercept and make a prediction. Hopefully this example will help you to make connections between the concepts we have presented.

Example

It costs $600 to purchase an iphone, plus $55 per month for unlimited use and data.

Write a linear equation that represents the cost, y,  of owning and using the iPhone for x amount of months. When you have written your equation, answer the following questions:

1. What is the total cost you’ve paid after owning and using your phone for 24 months?
2. If you have spent $2,580 since you purchased your phone, how many months have you used your phone?

iPhone

Show Solution

Read and Understand: We need to write a linear equation that represents the cost of owning and using an iPhone for any number of months.  We are to use y to represent cost, and x to represent the number of months we have used the phone.

Define and Translate: We will use the slope-intercept form of a line, [latex]y=mx+b[/latex], because we are given a starting cost and a monthly cost for use.  We will need to find the slope and the y-intercept.

Slope: in this case we don’t know two points, but we are given a rate in dollars for monthly use of the phone.  Our units are dollars per month because slope is [latex]\frac{\Delta{y}}{\Delta{x}}[/latex], and y is in dollars and x is in months. The slope will be [latex]\frac{55\text{ dollars }}{1\text{ month }}[/latex]. [latex]m=\frac{55}{1}=55[/latex]

Y-Intercept: the y-intercept is defined as a point [latex]\left(0,b\right)[/latex].  We want to know how much money we have spent, y, after 0 months.  We haven’t paid for service yet, but we have paid $600 for the phone. The y-intercept in this case is called an initial cost. [latex]b=600[/latex]

Write and Solve: Substitute the slope and intercept you defined into the slope=intercept equation.

[latex]\begin{array}{c}y=mx+b\\y=55x+600\end{array}[/latex]

Now we will answer the following questions:

1. What is the total cost you’ve paid after owning and using your phone for 24 months?

Since x represents the number of months you have used the phone, we can substitute x=24 into our equation.

[latex]\begin{array}{c}y=55x+600\\y=55\left(24\right)+600\\y=1320+600\\y=1920\end{array}[/latex]

Y represents the cost after x number of months, so in this scenario, after 24 months, you have spent $1920 to own and use an iPhone.

1. If you have spent $2,580 since you purchased your phone, how many months have you used your phone?

We know that y represents cost, and we are given a cost and asked to find the number of months related to having spent that much. We will substitute y=$2,580 into the equation, then use what we know about solving linear equations to isolate x:

 [latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=55x+600\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2580=55x+600\\\text{ subtract 600 from each side}\,\,\,\,\,\,\,\underline{-600}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-600}\\\text{}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1980=55x\\\text{}\\\text{ divide each side by 55 }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1980}{55}=\frac{55x}{55}\\\text{}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,36=x\end{array}[/latex]

If you have spent $2,580 then you have been using your iPhone for 36 months, or 3 years.

Summary

The slope-intercept form of a linear equation is written as [latex]y=mx+b[/latex], where m is the slope and b is the value of y at the y-intercept, which can be written as [latex](0,b)[/latex]. When you know the slope and the y-intercept of a line you can use the slope-intercept form to immediately write the equation of that line. The slope-intercept form can also help you to write the equation of a line when you know the slope and a point on the line or when you know two points on the line.

When lines in a plane are parallel (that is, they never cross), they have the same slope. When lines are perpendicular (that is, they cross at a 90° angle), their slopes are opposite reciprocals of each other. The product of their slopes will be [latex]-1[/latex], except in the case where one of the lines is vertical causing its slope to be undefined. You can use these relationships to find an equation of a line that goes through a particular point and is parallel or perpendicular to another line.