<meta http-equiv="refresh" content="1; url=/nojavascript/"> Distance Formula | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts Go to the latest version.

11.11: Distance Formula

Difficulty Level: At Grade Created by: CK-12
 0  0  0 Share To Groups
%
Best Score
Practice Distance Formula
Practice
Best Score
%
Practice Now

What if you were given the coordinates of two points like (6, 2) and (-3, 0). How could you determine how far apart these two points are? After completing this Concept, you'll be able to find the distance between any two points in the coordinate plane using the Distance Formula.

Watch This

CK-12 Foundation: The Distance Formula

Guidance

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.

Example A

Find the distance between points A = (1, 4) and B = (5, 2) .

Solution

Plot the two points on the coordinate plane.

In order to get from point A = (1, 4) to point B = (5, 2) , 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 A and B we find the value of the hypotenuse, d , using the Pythagorean Theorem.

d^2 &= 2^2+4^2=20\\d &= \sqrt{20}=2 \sqrt{5}=4.47

Example B

Find the distance between points C = (2, -1) and D = (-3, -4) .

Solution

We plot the two points on the graph above.

In order to get from point C to point D , we need to move 3 units down and 5 units to the left.

We find the distance from C to D by finding the length of d with the Pythagorean Theorem.

d^2 &= 3^2+5^2=34\\d &= \sqrt{34}=5.83

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 A=(x_1,y_1) and B=(x_2,y_2) .

Start by plotting the points on the coordinate plane:

In order to move from point A to point B in the coordinate plane, we move x_2 - x_1 units to the right and y_2 - y_1 units up.

We can find the length d by using the Pythagorean Theorem:

d^2=(x_1-x_2)^2+(y_1-y_2)^2

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

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

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 A to point B or from point B to point A , 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.

Example C

Find the distance between the following points.

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

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

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

Solution

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

a) d=\sqrt{(-3-4)^2+(5-(-2))^2}=\sqrt{49+49}=\sqrt{98}=7 \sqrt{2}

b) d=\sqrt{(12-19)^2+(16-21)^2}=\sqrt{49+25}=\sqrt{74}

c) d=\sqrt{(11.5+4.2)^2+(2.3+3.9)^2}=\sqrt{284.93}=16.88

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.

Example D

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

Solution

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.

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}

Notice that AB = AC , therefore triangle ABC is isosceles.

Example E

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?

Solution

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

The distance can be found with the distance formula:

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

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

Watch this video for help with the Examples above.

CK-12 Foundation: The Distance Formula

Vocabulary

  • The Distance Formula states that given any two points (x_1, y_1) and (x_2, y_2) , the distance between them is

d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.

Guided Practice

Find all points on the line y = 2 that are exactly 8 units away from the point (-3, 7).

Solution

Let’s make a sketch of the given situation.

Draw line segments from the point (-3, 7) to the line y = 2 .

Let k be the missing value of x we are seeking.

\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

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

Practice

Find the distance between the two points.

  1. (3, -4) and (6, 0)
  2. (-1, 0) and (4, 2)
  3. (-3, 2) and (6, 2)
  4. (0.5, -2.5) and (4, -4)
  5. (12, -10) and (0, -6)
  6. (-5, -3) and (-2, 11)
  7. (2.3, 4.5) and (-3.4, -5.2)
  8. Find all points having an x- coordinate of -4 whose distance from the point (4, 2) is 10.
  9. Find all points having a y- coordinate of 3 whose distance from the point (-2, 5) is 8.
  10. Find three points that are each 13 units away from the point (3, 2) but do not have an x- coordinate of 3 or a y- coordinate of 2.

Find the midpoint of the line segment joining the two points.

  1. Plot the points A = (1, 0), B = (6, 4), C = (9, -2) and D = (-6, -4), E = (-1, 0), F = (2, -6) . Prove that triangles ABC and DEF are congruent.
  2. Plot the points A = (4, -3), B = (3, 4), C = (-2, -1), D = (-1, -8). Show that ABCD is a rhombus (all sides are equal)
  3. Plot points A = (-5, 3), B = (6, 0), C = (5, 5). Find the length of each side. Show that ABC is a right triangle. Find its area.
  4. Find the area of the circle with center (-5, 4) and the point on the circle (3, 2).
  5. Michelle decides to ride her bike one day. First she rides her bike due south for 12 miles and then the direction of the bike trail changes and she rides in the 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.

Image Attributions

Description

Difficulty Level:

At Grade

Grades:

Date Created:

Oct 01, 2012

Last Modified:

Aug 21, 2014
Files can only be attached to the latest version of Modality

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
MAT.ALG.824.L.2
ShareThis Copy and Paste

Original text