# The Distance Formula

The **Distance Formula** is a useful tool in finding the *distance between two points* which can be arbitrarily represented as points \left( {{x_1},{y_1}} \right) and \left( {{x_2},{y_2}} \right). The distance formula itself is actually derived from the Pythagorean Theorem which is {a^2} + {b^2} = {c^2} where c is the longest side of a right triangle (also known as the hypotenuse) and a and b are the other shorter sides (known as the legs of a right triangle). That’s why we can claim that this formula is simply the Pythagorean Theorem in disguise.

If you want to see how the Distance Formula is derived from the Pythagorean Theorem, please check out my lesson on How to Derive the Distance Formula using the Pythagorean Theorem.

## Distance Formula as the Derivative of the Pythagorean Theorem

The distance d between two points

**and**

is calculated or computed using the following formula:

Below is an illustration showing that the Distance Formula is based on the Pythagorean Theorem where the distance d is the **hypotenuse** of a right triangle.

Observations:

a) The expression {x_2} - {x_1} is read as the “change in x“.

b) The expression {y_2} - {y_1} is read as the “change in y“.

### Examples of the Application of the Distance Formula

**Example 1:** Find the distance between the two points **(–3, 2)** and **(3, 5)**.

Label the parts of each point properly and substitute into the distance formula.

If we let \left( { - 3,2} \right) be the **first point** then it will take the subscript of 1, thus, {x_1} = - 3 and {y_1} = 2.

Similarly, if \left( {3,5} \right) be the **second point** it will have the subscript of 2, thus, {x_2} = 3 and {y_2} = 5.

Here is the calculation,

Therefore, the distance between two points **(–3, 2)** and **(3, 5) **is 3\sqrt 5 . This is how it looks on a graph.

**Example 2:** Find the distance between the points **(–1, –1)** and **(4, –5).**

If we assign \left( { - 1, - 1} \right) as our first point then

In the same manner, assigning \left( {4, - 5} \right) as our second point, we have

Plugging in the values of x and y, we get:

The two points and the distance between them which is \sqrt {41} can be shown on a graph just like the one below.

**Example 3:** Find the distance between the points **(–4, –3)** and **(4, 3)**.

Sometimes you may wonder if switching the points in calculating the distance can affect the final outcome.

Well, if you think about it, the formula is squaring the difference of the corresponding x and y values. That means it doesn’t matter if the change in x, also known as delta x, or the change in y, also known as delta y, is negative because when we eventually square it (raise to the 2nd power), the result always comes out to be positive.

Let’s “prove” that the answer is always the same by solving this problem two ways!

The first solution shows the usual way because we assign which point is the first and second based on the order in which they are given to us in the problem. In the second solution, we switch the points.

**Solution 1**:

**Solution 2**:

As you can see, both solutions arrived at the same answer or result which is the distance of 10, d = 10. Below is the visual solution to the problem.

**Example 4:** Find the radius of a circle with a diameter whose endpoints are **(–7, 1)** and **(1, 3)**.

Remember that the diameter of a circle is twice the length of its radius. If that’s the case, then the radius is half the length of the diameter.

Here’s the plan! Since we are given with the endpoints of the diameter, we can use the distance formula to find its length. Finally, we divide it by 2 to get the length of the radius, as required by the problem.

- Find the length of the diameter with endpoints \left( { - 7,1} \right) and \left( {1,3} \right).

- Solve for the radius by dividing the diameter by 2.

- The blue dots are the endpoints of diameter and the green dot is the center of the circle (calculated using the midpoint formula) located at \left( { - 3,2} \right).

##### Practice with Worksheets

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