### Video Transcript

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 coordinate
grid so we can see what’s going on. And it does just need to be a sketch. We don’t need
to measure it 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 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
25. And then adding them together gives me 𝑑 squared is equal to 34. Next step is to
square root both sides of this equation. So I have 𝑑 is equal to the square root of 34.
And if I evaluate this on my calculator, it gives me 𝑑 is equal to 5.83, 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 gonna call it 5.83 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 generalised
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 side squared,
plus 𝑦 two minus 𝑦 one squared, that’s the vertical side 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 an 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 25. 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 25. 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, it 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 26. Now I need to take the square root of both
sides. So 𝑑 is equal to the square root of 26. And if I evaluate that using a
calculator, I get 𝑑 is equal to 5.10 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 is 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 𝐵- 𝐵𝐶 squared is equal to 45. Then I
need to square root both sides. So 𝐵𝐶 is equal to the square root of 45. And then actually,
I can simplify this surd. Because what I need to remember is that 45 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 five.

Final step then is to calculate the area, so to multiply these two lengths together. So
I will have the area as root five times three root five. Now root five times root five
just gives me five. So I have five times three, which is 15. 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 15 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 an 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 they are perfectly happy just taking the Logical
approach of looking at the difference between the 𝑥-values, the 𝑦-values, and so on.