• Understand how to calculate the slope variable for a given line
• Understand how to calculate the y-intercept variable for a given line

Previously, we saw how a regression line can help us describe our relationship between an input variable like a movie budget and an output variable like the expected revenue from a movie with that budget. Let's take a look at that line again.

We also showed how we can describe a line with the formula $y = mx + b $, where $m$ is the slope and $b$ is the y-intercept of the line. One of the benefits of representing our line as a formula is that we then can calculate the $y$ value of our line for any input of $x$.

def y(x):
    return 3*x + 0 

y(0)

y(60000000)

We know what these $m$ and $b$ values represent. However, we still need to learn how to derive these values from an input line.

Say the following are a list of points along a line.

How do we calculate the slope $m$ given these points along the line?

This is the technique. We can determine the slope by taking any two points along the line and looking at the ** ratio of the vertical distance travelled to the horizontal distance travelled**. Rise over run.

Or, in math, it's:

$m = \frac{\Delta y}{\Delta x }$

The $\Delta$ is the capitalized version of the Greek letter Delta. Delta means change. So you can the read the above formula as $m$ equals change in $y$ divided by change in $x$.

Let's take another look of our graph and our line. Let's look at $x$ equal to zero and $x$ equal to 30 million. Plugging the numbers into our formula, we see that the slope between those two points is:

• $\Delta x$ = 30
• $\Delta y$ = 90
• $\frac{\Delta y}{\Delta x} = \frac{90}{30} = 3$

In other words, change in $x$ means our ending $x$ value minus our starting $x$ value and change in $y$ means our ending $y$ value minus our $y$ initial $y$ value .

Therefore, we can describe our $\Delta y$ and $\Delta x$ as the following:

• $\Delta y = y_1- y_0$
• $\Delta x = x_1 - x_0$

where $y_1$ is our ending point's $y$ value, $y_0$ is our initial point's $y$ value and $x_1$ and $x_0$ are our ending and initial $x$ values, respectively.

Altogether, we can say $m$ is the following:

given a beginning point $(x_0, y_0)$ and an ending point $(x_1, y_1)$ along any segment of a straight line, the slope of that line $m$ equals the following:

$$m = \frac{(y_1 - y_0)}{(x_1 - x_0)}$$

Now, let's apply this formula to our line. We can choose any two points along a straight line to calculate the slope of that line. So we now choose the second and third points in our table:

• our initial point (30, 90)
• our ending point of (60, 180)

Then plugging these coordinates into our formula, we have the following:

$m =\frac{(y_1 - y_0)}{(x_1 - x_0)} = \frac{(180 - 90)}{(60 - 30)} = 90/30 = 3$

So that is how we calculate the slope of a line.

Rise over run. Take any two points along that line and divide distance travelled vertically from the distance travelled horizontally. Change in $y$ divided by change in $x$.

Now that we know how to calculate the slope, let's turn our attention to calculating the y-intercept.

For example, look at the line below.

If you look at the far-left of the x-axis you will see that our $y$ value no longer is 0 when $x$ is 0. So we should calculate the value of our y-intercept, $b$. Here's what we can do.

First, let's figure out the slope of our line. Once again, we can choose any two points on the line to do this. So we choose the points (60, 208) and (30, 118). Plugging this into our formula for $m$ we have:

$m =\frac{(y_1 - y_0)}{(x_1 - x_0)} = \frac{(208 - 118)}{(60 - 30)} = 90/30 = 3$

Ok, now we plug in our value of $m$ into our formula:

$y = 3x + b $

Now to solve for $b$, we need to fill in values for $y$ and $x$.

It turns out we have lots of values for $y$ and $x$ to fill in. Any $y$ and $x$ value along the line can be used. For example, we can see from our line that when x = 30, y = 118. Let's plug that into our formula and solve for $b$:

$ 118 = 3 * 30 + b $

$ 118 = 90 + b $

$ 28 = b $

Solving for $b$, we see that $b$ = 28. So now we have filled in our $m$ and $b$ variables for our line, and can describe our blue line above by the formula $y = 3x + 28 $. Or in code it looks like:

def y(x):
    return 3*x + 28

Let's see how well we did by providing a value of $x$, and seeing if the $y$ value lines up to the $y$ value of the line in our chart.

y(20)

When plugging an $x$ value of 20 million into our formula, we see that $y$ equals 88 million. Let's look at our graph above and compare this result to the $y$ value where $x$ is 20 million. It seems we did a good job of getting the slope by calculating $m$ and using an $x$ and $y$ value pair to solve for the y-intercept.

In this lesson, we saw how to calculate the slope and y-intercept variables that describe a line. We can take any two points along the line to calculate our slope variable. This is because given two points along the straight line, we can divide the change in $y$ over those two points by the change in $x$ over those two points to get the slope. Then we can take that slope ($m$), pick a point along the line, plug that point's $x$ value and $y$ value into the $y = mx + b$ formula, and finally use algebra to solve for $b$ to discover the y-intercept.