### The Distance Formula

In the last section, we saw how to use the Pythagorean Theorem to find lengths. In this section, you’ll learn how to use the Pythagorean Theorem to find the distance between two coordinate points.

#### Finding the Distance Between Two Points

1. Find the distance between points \begin{align*}A = (1, 4)\end{align*} and \begin{align*}B = (5, 2)\end{align*}.

Plot the two points on the coordinate plane.

In order to get from point \begin{align*}A = (1, 4)\end{align*} to point \begin{align*}B = (5, 2)\end{align*}, we need to move 4 units to the right and 2 units down. These lines make the legs of a right triangle.

To find the distance between \begin{align*}A\end{align*} and \begin{align*}B\end{align*} we find the value of the hypotenuse, \begin{align*}d\end{align*}, using the Pythagorean Theorem.

\begin{align*}d^2 &= 2^2+4^2=20\\ d &= \sqrt{20}=2 \sqrt{5}=4.47\end{align*}

2. Find the distance between points \begin{align*}C = (2, -1)\end{align*} and \begin{align*}D = (-3, -4)\end{align*}.

We plot the two points on the graph above.

In order to get from point \begin{align*}C\end{align*} to point \begin{align*}D\end{align*}, we need to move 3 units down and 5 units to the left.

We find the distance from \begin{align*}C\end{align*} to \begin{align*}D\end{align*} by finding the length of \begin{align*}d\end{align*} with the Pythagorean Theorem.

\begin{align*}d^2 &= 3^2+5^2=34\\ d &= \sqrt{34}=5.83\end{align*}

**The Distance Formula**

The procedure we just used can be generalized by using the Pythagorean Theorem to derive a formula for the distance between any two points on the coordinate plane.

Let’s find the distance between two general points \begin{align*}A=(x_1,y_1)\end{align*} and \begin{align*}B=(x_2,y_2)\end{align*}.

Start by plotting the points on the coordinate plane:

In order to move from point \begin{align*}A\end{align*} to point \begin{align*}B\end{align*} in the coordinate plane, we move \begin{align*}x_2 - x_1\end{align*} units to the right and \begin{align*}y_2 - y_1\end{align*} units up.

We can find the length \begin{align*}d\end{align*} by using the Pythagorean Theorem:

\begin{align*}d^2=(x_1-x_2)^2+(y_1-y_2)^2\end{align*}

Therefore, \begin{align*}d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\end{align*}. This is called the **Distance Formula.** More formally:

Given any two points \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2)\end{align*}, the distance between them is \begin{align*}d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.\end{align*}

We can use this formula to find the distance between any two points on the coordinate plane. Notice that the distance is the same whether you are going from point \begin{align*}A\end{align*} to point \begin{align*}B\end{align*} or from point \begin{align*}B\end{align*} to point \begin{align*}A\end{align*}, so it does not matter which order you plug the points into the distance formula.

Let’s now apply the distance formula to the following examples.

#### Using the Distance Formula

Plug the values of the two points into the distance formula. Be sure to simplify if possible.

a) (-3, 5) and (4, -2)

\begin{align*}d=\sqrt{(-3-4)^2+(5-(-2))^2}=\sqrt{49+49}=\sqrt{98}=7 \sqrt{2}\end{align*}

b) (12, 16) and (19, 21)

\begin{align*}d=\sqrt{(12-19)^2+(16-21)^2}=\sqrt{49+25}=\sqrt{74}\end{align*}

c) (11.5, 2.3) and (-4.2, -3.9)

\begin{align*}d=\sqrt{(11.5+4.2)^2+(2.3+3.9)^2}=\sqrt{284.93}=16.88\end{align*}

**Applications Using Distance and Midpoint Formulas**

The distance and midpoint formula are useful in geometry situations where we want to find the distance between two points or the point halfway between two points.

#### Proving that a Triangle is Isosceles

Plot the points *\begin{align*}A = ( 4, -2), B = (5, 5)\end{align*}, and \begin{align*}C = (-1, 3)\end{align*} and connect them to make a triangle. Show that the triangle is isosceles.*

Let’s start by plotting the three points on the coordinate plane and making a triangle:

We use the distance formula three times to find the lengths of the three sides of the triangle.

\begin{align*}AB &= \sqrt{(4-5)^2+(-2-5)^2}=\sqrt{(-1)^2+(-7)^2}=\sqrt{50}=5 \sqrt{2}\\ BC &= \sqrt{(5+1)^2+(5-3)^2}=\sqrt{(6)^2+(2)^2}=\sqrt{40}=2 \sqrt{10}\\ AC &= \sqrt{(4+1)^2+(-2-3)^2}=\sqrt{(5)^2+(-5)^2}=\sqrt{50}=5 \sqrt{2}\end{align*}

Notice that \begin{align*}AB = AC\end{align*}, therefore triangle \begin{align*}ABC\end{align*} is isosceles.

#### Real-World Application: Walking Speed

At 8 AM one day, Amir decides to walk in a straight line on the beach. After two hours of making no turns and traveling at a steady rate, Amir is two miles east and four miles north of his starting point. How far did Amir walk and what was his walking speed?

Let’s start by plotting Amir’s route on a coordinate graph. We can place his starting point at the origin: \begin{align*}A = (0, 0)\end{align*}. Then his ending point will be at \begin{align*}B = (2, 4)\end{align*}.

The distance can be found with the distance formula:

\begin{align*}d &= \sqrt{(2-0)^2+(4-0)^2}=\sqrt{(2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}\\ d &= \underline{\underline{4.47 \ miles}}\end{align*}

Since Amir walked 4.47 miles in 2 hours, his speed is \begin{align*}s=\frac{4.47 \ miles}{2 \ hours}=\underline{\underline{2.24 \ mi/h}}\end{align*}.

### Example

#### Example 1

Find all points on the line \begin{align*}y = 2\end{align*} that are exactly 8 units away from the point (-3, 7).

Let’s make a sketch of the given situation.

Draw line segments from the point (-3, 7) to the line \begin{align*}y = 2\end{align*}.

Let \begin{align*}k\end{align*} be the missing value of \begin{align*}x\end{align*} we are seeking.

\begin{align*}\text{Let's use the distance formula:} && 8& =\sqrt{(-3-k)^2+(7-2)^2}\\ \text{Square both sides of the equation:} && 64& =(-3-k)^2+25\\ \text{Therefore:} && 0& =9+6k+k^2-39 \ \text{or} \ 0=k^2+6k-30\\ \text{Use the quadratic formula:} && k& =\frac{-6 \pm \sqrt{36 + 120}}{2}=\frac{-6 \pm \sqrt{156}}{2}\\ \text{Therefore:} && k& =3.24 \ \text{or} \ k=-9.24\end{align*}

The points are **(-9.24, 2) and (3.24, 2).**

### Review

Find the distance between the two points.

- (3, -4) and (6, 0)
- (-1, 0) and (4, 2)
- (-3, 2) and (6, 2)
- (0.5, -2.5) and (4, -4)
- (12, -10) and (0, -6)
- (-5, -3) and (-2, 11)
- (2.3, 4.5) and (-3.4, -5.2)
- Find all points having an \begin{align*}x-\end{align*}coordinate of -4 whose distance from the point (4, 2) is 10.
- Find all points having a \begin{align*}y-\end{align*}coordinate of 3 whose distance from the point (-2, 5) is 8.
- Find three points that are each 13 units away from the point (3, 2) but do
*not*have an \begin{align*}x-\end{align*}coordinate of 3 or a \begin{align*}y-\end{align*}coordinate of 2.

- Plot the points \begin{align*}A = (1, 0), B = (6, 4), C = (9, -2)\end{align*} and \begin{align*}D = (-6, -4), E = (-1, 0), F = (2, -6)\end{align*}. Prove that triangles \begin{align*}ABC\end{align*} and \begin{align*}DEF\end{align*} are congruent.
- Plot the points \begin{align*}A = (4, -3), B = (3, 4), C = (-2, -1), D = (-1, -8).\end{align*} Show that \begin{align*}ABCD\end{align*} is a rhombus (all sides are equal)
- Plot points \begin{align*}A = (-5, 3), B = (6, 0), C = (5, 5).\end{align*} Find the length of each side. Show that \begin{align*}ABC\end{align*} is a right triangle. Find its area.
- Find the area of the circle with center (-5, 4) and a point on the circle (3, 2).
- Michelle decides to ride her bike one day. First she rides her bike due south for 12 miles and then she rides in a new direction for a while longer. When she stops Michelle is 2 miles south and 10 miles west from her starting point. Find the total distance that Michelle covered from her starting point.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 11.11.