# Video: Graphing Experimental Data

What is the value of the gradient of the line of best fit for the points plotted on the graph shown?

05:03

### Video Transcript

What is the value of the gradient of the line of best fit for the points plotted on the graph shown?

So in this question, we’re being asked to find the value of the gradient of the line of best fit for the points plotted on the graph shown. Now, that’s a lot of information in a very small sentence. But basically, what we’re being told is that we’ve got some data which consists of these points here: one, two, three, four, five, six points. And this blue line has been drawn as the line of best fit for these points. What we need to do is to find out the gradient of this line of best fit.

Now, the data points are only there on the graph to show us that we have actually drawn a line of best fit. But in reality, if we want to find a gradient of the line of best fit, then we don’t need to know what the data points are because we already have the line. So we can ignore the data points for now. Anyway, how do we find the gradient of this line which happens to be a line of best fit? Well, we can recall that the gradient is defined as rise over run or if we call the horizontal axis the 𝑥-axis and the vertical axis the 𝑦-axis, the gradient is also defined as the change in 𝑦 divided by the change in 𝑥.

Now both of these are telling us exactly the same thing just using different words. In order to calculate the gradient of the line, we first need to choose two points randomly on the line itself. Now to make life easier for ourselves, we’ll choose these two points here. Firstly, the line goes through the origin. So we choose the origin as one point. And secondly, it goes through this point here.

Now, we can write down the coordinates of these two points. Well, the coordinates of the origin of course are zero, zero. We’re going zero along 𝑥 and zero along 𝑦. And the coordinates of the other point are? Well, we’ve gone five along in the 𝑥-direction, so five, and we’ve gone six up. So five, six are our coordinates because just to be clear we’ve gone five along in this direction and six up in this direction.

So at this point, we’ve got the coordinates of two points on the line, which means we can calculate the gradient because we can work out the rise and the run or we can work out the change in 𝑦 and the change in 𝑥.

Let’s start by calculating the rise over run. Well, the rise is defined as this distance here. It’s the vertical distance between our two points. And in our case, the two points are at the origin and that five, six. So that’s the rise. That’s this value here. The run however is this distance here. It’s the horizontal distance between the two points and once again that distance is the run.

So now, what are the rise and the run? Well, the rise — the vertical distance — is six because in order to go from zero, zero to five, six, we have to go up by six units. And similarly, in order to go from zero, zero to five, six, we have to go across five units. Therefore, if we wanna calculate our rise over run, well we’ve said that the rise is six units and the run is five units, so our gradient is going to be equal to six divided by five. And since we don’t have any units on the axes, we can just say that the gradient has some arbitrary units. Anyway, so the gradient which is rise over run or six divided by five is equal to 1.2 because that’s what’s six divided by five evaluates to.

Now, we can show this using the change in 𝑦 divided by the change in 𝑥 method: firstly, the change in 𝑦. The change in 𝑦 is simply the difference between the 𝑦-coordinates of the two points that we’ve chosen on the line. In other words, it’s this coordinate which is six minus this coordinate here which is zero because those are the two 𝑦-coordinates. And so we find that the change in 𝑦 is six minus zero, which simplifies to just six.

Similarly, for the change in 𝑥, we’ve got this coordinate here minus this coordinate here. So that’s five minus zero, which yet again simplifies to just five. And once again, our change in 𝑦 divided by our change in 𝑥 is six divided by five, which happens to be 1.2.

So as we expect, we get the same answer using both methods. And it’s worth noting by the way that if we had a more accurate set of axes, it’s important to note by the way that we would get the same gradient regardless of which two points on the line we chose. So if we chosen this point here and this point here, we would have still found the rise and run or change in 𝑦 and change in 𝑥 such that the gradient would still be 1.2.

But because we don’t have a very accurate set of axes, we just picked the points that made life easiest for ourselves. And that happened to be the origin and this point here. And the reason is because they corresponded to nice integer numbers on the axes. It’s also worth reiterating that we wanted to find the gradient of the line. Therefore, we did not pick two points that were part of our dataset. We didn’t pick these two for example. Instead, we picked two points on the line because we wanted to find the gradient of the line.

The data points — the random ones that were off the line — were just there to show us that the line of best fit was fairly accurately drawn. But anyway, at this point, we now know our answer. The gradient of the line of best fit is 1.2.