In this video, we are going to look at a particular application of the Pythagorean theorem, which is finding the distance between two points on a coordinate grid. And we’ll look at this both in two dimensions and also in three dimensions. So let’s start off with an example in two dimensions, and the question we’ve got is to find the distance between the points with coordinates negative three, one and two, four.
So in order to start with this question, it’s best to do a sketch of the co-ordinate grid so we can see what’s going on. And it does just need to be a sketch; we don’t need to measure accurately; we don’t need squared paper, just a sketch of a two dimensional coordinate grid with these points marked on it. So here is my sketch of that coordinate grid with the approximate positions of the points negative three, one and two, four. So if I must find the distance between these two points, then I’m looking for the direct distance if I join them up with a straight line. So I’m looking to calculate this direct distance here between those two points. Now the Pythagorean theorem is all about right-angled triangles, so I need to create a right-angled triangle. And what I can do is, either above or below this line, I can sketch in — This is a little right-angled triangle here, so that then I have the right-angled triangle that I can use with the Pythagorean theorem. Now I’m looking to calculate this distance so I’ll give it the letter 𝑑. And what I need to think about are what are the lengths of these other two sides of the triangle. So let’s look at the horizontal distance first of all. This horizontal distance — Well the only thing that’s changing is the 𝑥-coordinate, and it’s changing from negative three to two, which means this distance here, the horizontal part of that triangle, must be five units. So I can fill that in. Now if I look at the vertical side of the triangle, well here the only thing that’s changing is the 𝑦-coordinate. And it’s changing from one here to four here, which means this side of the triangle must be equal to three units. So now I have the right setup for the Pythagorean theorem. I know two sides of the triangle, and I want to calculate the third, in this case the hypotenuse.
So a reminder of the Pythagorean theorem, it tells us that 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared where 𝑎 and 𝑏 represent the two shorter sides of a right-angled triangle and 𝑐 represents the hypotenuse. So the first step then is just to write down what the Pythagorean theorem tells me, specifically for this triangle here. So if I write that down, I will have 𝑑 squared, the hypotenuse squared, is equal to three squared plus five squared. Then I can replace both of those with their values, nine and twenty-five. And then adding them together gives me 𝑑 squared is equal to thirty-four. Next step is to square root both sides of this equation. So I have 𝑑 is equal to the square root of thirty-four, and if I evaluate this on my calculator it gives me 𝑑 is equal to five point eight three to three significant figures. Now units for this, we haven’t been told that it’s a centimetre square grid, so we can’t assume units are centimetres. The units are just going to be general distance units or general length units. So I’m just going to call it five point eight three units and, as I said, that was rounded to three significant figures.
Now let’s look at how we can generalise this. So if we can come up with a generalized distance formula that we can use to calculate the distance between any two points, and I’ve called them 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two to represent general points on a coordinate grid. Now as before, we’ll start with a sketch. We don’t know anything about 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, but we’ll just assume arbitrarily that they form a line that looks something like this. So we have 𝑥 one, 𝑦 one down here and we have 𝑥 two, 𝑦 two here. And we’re looking to calculate the distance between those two points. So as before, I would need to fill in the little right-angled triangle below the line. Now I need to work out the lengths of the two sides of this triangle. So let’s look at the 𝑥-coordinate first. Now it’s changing form 𝑥 one at this point here to 𝑥 two at this point here. So the length of that line is gonna be the difference between those two 𝑥-values. So it’s going to be 𝑥 two minus 𝑥 one. Now if I look at the length of the vertical line, I’m gonna have a similar type of thing. So on the vertical line, the 𝑦-coordinate is changing, and it’s changing from 𝑦 one at this point here to 𝑦 two at this point here. So the length of that vertical line is gonna be the difference between those two 𝑦-values; it’s going to be 𝑦 two minus 𝑦 one. So that gives me generalised formulae for the lengths of the two sides of this triangle.
Right! Now I can write down what the Pythagorean theorem tells me in terms of 𝑑 and 𝑥 one, 𝑥 two, 𝑦 one, and 𝑦 two. So what I’m gonna have, 𝑑 squared, the hypotenuse squared is equal to 𝑥 two minus 𝑥 one squared, that’s the horizontal sides squared, plus 𝑦 two minus 𝑦 one squared; that’s the vertical sides squared. The final step in deriving this generalised formula is I want to know 𝑑, not 𝑑 squared. So I need to take the square root of both sides of this equation. And if I do that, I get this general formula here: 𝑑 is equal to the square root of 𝑥 two minus 𝑥 one all squared plus 𝑦 two minus 𝑦 one all squared. And that is a generalised distance formula for calculating the distance between two points 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two. Now it doesn’t actually matter in the context of example which point we consider to be 𝑥 one, 𝑦 one and which we consider to be 𝑥 two, 𝑦 two, because what you’re doing is you’re finding the difference between the 𝑥-values and the difference between the 𝑦-values and squaring it. And if you do that one way round, you will get for example a difference of five and square it to twenty-five. If you do it the other way around you’ll get a difference of negative five, but when you Square it you will still get positive twenty-five. So you can think of these two points in either order. Now this generalised formula is useful because it gives us a formula that will always work and we can plug any numbers into it, but in the previous example all we did was take a purely logical approach to answering the question. And personally I sometimes find actually it’s easier just to take a logical approach rather than using this distance formula.
Okay now let’s look at an example in three dimensions. So you’ll have seen before that the Pythagorean theorem can be extended into three dimensions. And when we’re working in three dimensions, we have the formula 𝑎 squared plus 𝑏 squared plus 𝑐 squared is equal to 𝑑 squared. So let’s look at applying this in this case. We want to work out the distance between these two points, so we want 𝑑 squared. Now first of all, let’s look at the difference between the 𝑥-coordinates; the 𝑥-coordinates change from two to negative one, which is a change of negative three. Now as mentioned on the previous example, doesn’t actually matter whether I call it three or negative three because when I square it I’m gonna get the same result. So I’ll just think of it as three. The 𝑦 value changes from zero to four, so we’ve got plus four squared, and then the 𝑧-value in this case, in the three-dimensional coordinate grid, changes from five to four, so it’s a difference of one. And so we’ll have one squared. So there’s my statement of the Pythagorean theorem in three dimensions for this particular question. The next step is to work out three squared, four squared, and one squared. And then if I add them all together, I get 𝑑 squared is equal to twenty-six. Now I need to take the square root of both sides. So 𝑑 is equal to the square root of twenty-six. And if I evaluate that using a calculator, I get 𝑑 is equal to five point one zero units, length units or distance units, and that value has been rounded to three significant figures. So just a reminder of what we did here, we looked at the difference between the 𝑥-coordinates, which was three, the difference between the 𝑦-coordinates, which was four, and the difference between the 𝑧-coordinates, which was one, and then we used the three-dimensional version of the Pythagorean theorem in order to calculate the distance between these two points in three-dimensional space.
Finally, let’s look at an application of this. So we have the question the vertices of a rectangle 𝐴𝐵𝐶𝐷 are these four points here. Find the area of the rectangle. So to find the area of the rectangle, we need to know the lengths of its two sides. So we’re going to be using the Pythagorean theorem twice in order to calculate two lengths. Now as always, let’s just start off with a sketch so we can picture what’s happening here. So here we have a sketch of that coordinate grid with the points 𝐴, 𝐵, 𝐶 and 𝐷 marked on in their approximate positions, and you can see that by joining them up we form this rectangle. So in order to calculate the area of this rectangle, I need to work out the lengths of its two sides and then multiply them together. So I’m gonna do the area of this rectangle; I’m gonna find the length of 𝐴𝐵, and I’m gonna multiply it by 𝐵𝐶, but equally I could have done 𝐶𝐷 multiplied by 𝐴𝐷 or whichever combination I particularly wanted to do. So let’s find the length of 𝐴𝐵 first. so I’m interested in the points three, three and two, one in order to do this. So let’s work out this length using the Pythagorean theorem. So 𝐴𝐵 squared, the 𝑥 coordinates — Well the difference between those, it goes from two to three. So that’s a difference of one, so one squared. And then the difference between the 𝑦-coordinates, it goes from one to three, difference of two, two squared. So there is a statement of the Pythagorean theorem to calculate 𝐴𝐵. So the next two stages, work out what one squared and two squared are and then add them together, and then I need to square root both sides. And I’ll leave it as 𝐴𝐵 is equal to the square root of five for now. So we’ve got one length worked out. Now I need to do the same thing for 𝐵𝐶. So 𝐵𝐶 squared, if I look at the 𝑥-coordinate, it’s changing from two to negative four so that’s negative six. But remember, it doesn’t matter whether I call it positive or negative, so I’ll just keep it as six squared. If I look at the 𝑦-coordinate, it’s changing from one to four. So there’s a difference of three there, so three squared. And there’s our statement of the Pythagorean theorem to calculate 𝐵𝐶. So then I work out what six squared and three squared are, and then I add them together and I get 𝐵𝐶 squared is equal to forty-five. Then I need to square root both sides. So 𝐵𝐶 is equal to the square root of forty-five. And then I check and simplify this surd because what I need to remember is that forty-five is equal to nine times five. And because nine is a square number, I can bring that square root of nine outside the front, and it will simplify as a surd to 𝐵𝐶 is equal to three root five. So there I have the lengths of my two sides: 𝐴𝐵 equals root five, 𝐵𝐶 equals three root three.
Final step then is to calculate the area, so to multiply these two lengths together. So I would have the area is root five times three root five. Now root five times root five just gives me five. So I have five times three, which is fifteen. Now units for this — Well it’s an area, so it needs to be square units. We don’t know whether it’s square centimetres or square millimetres, so we’ll just call it fifteen square units for the area. So in this question, it involved applying the Pythagorean theorem twice to find the distance between two different sets of points and then combining them using what we know about areas of rectangles. So there you have a summary of how to use the Pythagorean theorem to calculate the distance between two points, and we saw how to do this in two dimensions. We saw also how to do it in three dimensions and then in application to finding the area of a rectangle. We saw also how to generalise, to come up with that distance formula. And you may find it helpful to use that if you like to just substitute into a formula or you may find that you’re perfectly happy just taking the Logical approach of looking at the difference between the 𝑥-values, the 𝑦-values, and so on.