Distance between two points on a graph

It is sometimes useful to find the distance between two points on a graph. For example, we might need to find the distance between the points A and B on this graph:

We can find the distance between the two points using the coordinates of the points. A is at (2, 1) and B is at (6, 4). In this article, we will use Pythagoras' theorem to find the length.

Horizontal and vertical distances

Before looking at the general case, we will take a look at a couple of simpler cases, shown here:

First, we will find the distance from P to Q. P is at (1, 3) and Q is at (4, 3). The line PQ is horizontal, because the two points P and Q have the same y value. The distance between P and Q is the horizontal distance, and it can be found by subtracting the x value of P from the x value of Q:

Next, we will find the distance from R to S. R is at (6, 7) and S is at (6, 2). The line RS is vertical because the two points R and S have the same x value. The distance between R and S is the vertical distance, and it can be found by subtracting the y value of R from the y value of S:

Notice that this gives a negative distance. That is because R is above S so we have to move in the negative y direction to get from R to S.

When we talk about the distance between two points, we don't usually care about the direction. For example, the distance between London and Edinburgh is 332 miles, and the distance between Edinburgh and London is also 332 miles. We don't say that one of those distances is minus 332 miles!

So we would normally ignore the sign and say that the distance between R and S is 5. We will see another explanation for this in the next section.

Diagonal distances

In this case, the line AB is neither horizontal nor vertical:

If we draw a straight line between A and B, that line has length c. That is the distance between A and B. How can we find the distance? Well, we can construct a right-angled triangle, and use Pythagoras' theorem.

Point C is horizontally in line with A and vertically in line with B. So we can find the coordinates of C, because it has the same y coordinate as A and the same x coordinate as B:

We cannot find c directly, but we can find:

• a (the vertical distance between B and C), which is the y component of C minus the y component of B.
• and b (the horizontal distance between A and C), which is the x component of C minus the x component of A.

Since ABC forms a right-angled triangle, and we know a and b, we can find c using Pythagoras' theorem, which tells us that:

We can solve for c like this:

Horizontal and vertical distances are special cases

We can apply the Pythagoras method to horizontal and vertical lines, and we will see that they are just special cases of the general case. Here is the diagram again with the line lengths labelled:

The line PQ can be thought of as a "triangle" with a width b but a height of zero. We can find b as before:

We can find the distance between the points using Pythagoras, but with a set to zero:

As we expect, the distance c is equal to 3. But c will always be positive even for a negative b.

The line RS can also be thought of as a "triangle", but this time with height a but a width of zero. We can find a as before:

Again, we use Pythagoras, but with b set to zero:

So now, even though a was negative, the distance between the points, c, is positive.

Example 1

As an example, we will find the length of the line AB from the triangle example above:

The points on the triangle are:

We can calculate a and b:

The value of a is negative, but as we saw earlier that does not matter as we will be squaring it anyway. We use Pythagoras to find c squared:

We find c by taking the square root:

So the distance between A and B is 5. This is a 3-4-5 Pythagorean triple.

Example 2

In this next example, we will find the length of the line DE below:

The points on the triangle are:

The difference in this case is that the points E and F are on the negative side of the y axis. That is no problem, but we need to be consistent with the signs:

Pay particular attention to the way b is calculated. The x component of F is negative (-4) and the x component of D is positive (8), so to find b we calculate (-4) - 8 which is -12.

So this time a and b are both negative, but again the squares are both positive. We use Pythagoras to find c:

So the distance between D and E is 13. This is a 5-12-13 Pythagorean triple.

Example 3

Sometimes the question will just show the points, not the entire triangle. In this final example, we will find the length of the line UV below:

We need to redraw the points, on axes, showing the third point (which we will call W):

The points on the triangle are:

As before, we need to be consistent with the signs when we do the calculation. Here are the values of a and b

We use Pythagoras to find c:

This time the triangle is not a Pythagorean triple, so the length can only be expressed exactly as a surd. The distance between U and V to 3 decimal places is 5.831.

See also

• Pythagoras' theorem
• Pythagorean triples
• Pythagoras' theorem proof
• Einstein's proof of Pythagoras' theorem