assignment #9 finding midpoints and distance

11.1 Distance and Midpoint Formulas; Circles

Learning objectives.

By the end of this section, you will be able to:

Use the Distance Formula

Use the Midpoint Formula

Write the equation of a circle in standard form
Graph a circle

Be Prepared 11.1

Before you get started, take this readiness quiz.

Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches. If you missed this problem, review Example 2.34 .

Be Prepared 11.2

Factor: x 2 − 18 x + 81 . x 2 − 18 x + 81 . If you missed this problem, review Example 6.24 .

Be Prepared 11.3

Solve by completing the square: x 2 − 12 x − 12 = 0 . x 2 − 12 x − 12 = 0 . If you missed this problem, review Example 9.22 .

In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

There are four conics—the circle , parabola , ellipse , and hyperbola . The next figure shows how the plane intersecting the double cone results in each curve.

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions . We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Example 11.1

Use the rectangular coordinate system to find the distance between the points ( 6 , 4 ) ( 6 , 4 ) and ( 2 , 1 ) . ( 2 , 1 ) .

Try It 11.1

Use the rectangular coordinate system to find the distance between the points ( 6 , 1 ) ( 6 , 1 ) and ( 2 , −2 ) . ( 2 , −2 ) .

Try It 11.2

Use the rectangular coordinate system to find the distance between the points ( 5 , 3 ) ( 5 , 3 ) and ( −3 , −3 ) . ( −3 , −3 ) .

The method we used in the last example leads us to the formula to find the distance between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) .

When we found the length of the horizontal leg we subtracted 6 − 2 6 − 2 which is x 2 − x 1 . x 2 − x 1 .

When we found the length of the vertical leg we subtracted 4 − 1 4 − 1 which is y 2 − y 1 . y 2 − y 1 .

If the triangle had been in a different position, we may have subtracted x 1 − x 2 x 1 − x 2 or y 1 − y 2 . y 1 − y 2 . The expressions x 2 − x 1 x 2 − x 1 and x 1 − x 2 x 1 − x 2 vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say | x 2 − x 1 | | x 2 − x 1 | and | y 2 − y 1 | . | y 2 − y 1 | .

In the Pythagorean Theorem, we substitute the general expressions | x 2 − x 1 | | x 2 − x 1 | and | y 2 − y 1 | | y 2 − y 1 | rather than the numbers.

This is the Distance Formula we use to find the distance d between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) .

Distance Formula

The distance d between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) ( x 2 , y 2 ) is

Example 11.2

Use the Distance Formula to find the distance between the points ( −5 , −3 ) ( −5 , −3 ) and ( 7 , 2 ) . ( 7 , 2 ) .

Try It 11.3

Use the Distance Formula to find the distance between the points ( −4 , −5 ) ( −4 , −5 ) and ( 5 , 7 ) . ( 5 , 7 ) .

Try It 11.4

Use the Distance Formula to find the distance between the points ( −2 , −5 ) ( −2 , −5 ) and ( −14 , −10 ) . ( −14 , −10 ) .

Example 11.3

Use the Distance Formula to find the distance between the points ( 10 , −4 ) ( 10 , −4 ) and ( −1 , 5 ) . ( −1 , 5 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Try It 11.5

Use the Distance Formula to find the distance between the points ( −4 , −5 ) ( −4 , −5 ) and ( 3 , 4 ) . ( 3 , 4 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Try It 11.6

Use the Distance Formula to find the distance between the points ( −2 , −5 ) ( −2 , −5 ) and ( −3 , −4 ) . ( −3 , −4 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the x -coordinates and the average of the y -coordinates of the endpoints.

Midpoint Formula

The midpoint of the line segment whose endpoints are the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) ( x 2 , y 2 ) is

To find the midpoint of a line segment, we find the average of the x -coordinates and the average of the y -coordinates of the endpoints.

Example 11.4

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −5 , −4 ) ( −5 , −4 ) and ( 7 , 2 ) . ( 7 , 2 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Try It 11.7

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −3 , −5 ) ( −3 , −5 ) and ( 5 , 7 ) . ( 5 , 7 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Try It 11.8

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −2 , −5 ) ( −2 , −5 ) and ( 6 , −1 ) . ( 6 , −1 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Both the Distance Formula and the Midpoint Formula depend on two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) . It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, it may be easier to remember the formulas.

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, ( h , k ) , ( h , k ) , and the fixed distance is called the radius , r , of the circle.

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center , ( h , k ) , ( h , k ) , and the fixed distance is called the radius , r , of the circle.

This is the standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r .

Standard Form of the Equation a Circle

The standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r , is

Example 11.5

Write the standard form of the equation of the circle with radius 3 and center ( 0 , 0 ) . ( 0 , 0 ) .

Try It 11.9

Write the standard form of the equation of the circle with a radius of 6 and center ( 0 , 0 ) . ( 0 , 0 ) .

Try It 11.10

Write the standard form of the equation of the circle with a radius of 8 and center ( 0 , 0 ) . ( 0 , 0 ) .

In the last example, the center was ( 0 , 0 ) . ( 0 , 0 ) . Notice what happened to the equation. Whenever the center is ( 0 , 0 ) , ( 0 , 0 ) , the standard form becomes x 2 + y 2 = r 2 . x 2 + y 2 = r 2 .

Example 11.6

Write the standard form of the equation of the circle with radius 2 and center ( −1 , 3 ) . ( −1 , 3 ) .

Try It 11.11

Write the standard form of the equation of the circle with a radius of 7 and center ( 2 , −4 ) . ( 2 , −4 ) .

Try It 11.12

Write the standard form of the equation of the circle with a radius of 9 and center ( −3 , −5 ) . ( −3 , −5 ) .

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Example 11.7

Write the standard form of the equation of the circle with center ( 2 , 4 ) ( 2 , 4 ) that also contains the point ( −2 , 1 ) . ( −2 , 1 ) .

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center ( 2 , 4 ) ( 2 , 4 ) and point ( −2 , 1 ) ( −2 , 1 )

Now that we know the radius, r = 5 , r = 5 , and the center, ( 2 , 4 ) , ( 2 , 4 ) , we can use the standard form of the equation of a circle to find the equation.

Try It 11.13

Write the standard form of the equation of the circle with center ( 2 , 1 ) ( 2 , 1 ) that also contains the point ( −2 , −2 ) . ( −2 , −2 ) .

Try It 11.14

Write the standard form of the equation of the circle with center ( 7 , 1 ) ( 7 , 1 ) that also contains the point ( −1 , −5 ) . ( −1 , −5 ) .

Graph a Circle

Any equation of the form ( x − h ) 2 + ( y − k ) 2 = r 2 ( x − h ) 2 + ( y − k ) 2 = r 2 is the standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r. We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from x and y . In the next example, the equation has x + 2 , x + 2 , so we need to rewrite the addition as subtraction of a negative.

Example 11.8

Find the center and radius, then graph the circle: ( x + 2 ) 2 + ( y − 1 ) 2 = 9 . ( x + 2 ) 2 + ( y − 1 ) 2 = 9 .

Try It 11.15

ⓐ Find the center and radius, then ⓑ graph the circle: ( x − 3 ) 2 + ( y + 4 ) 2 = 4 . ( x − 3 ) 2 + ( y + 4 ) 2 = 4 .

Try It 11.16

ⓐ Find the center and radius, then ⓑ graph the circle: ( x − 3 ) 2 + ( y − 1 ) 2 = 16 . ( x − 3 ) 2 + ( y − 1 ) 2 = 16 .

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of x 2 , y 2 x 2 , y 2 to be one.

Example 11.9

Find the center and radius and then graph the circle, 4 x 2 + 4 y 2 = 64 . 4 x 2 + 4 y 2 = 64 .

Try It 11.17

ⓐ Find the center and radius, then ⓑ graph the circle: 3 x 2 + 3 y 2 = 27 3 x 2 + 3 y 2 = 27

Try It 11.18

ⓐ Find the center and radius, then ⓑ graph the circle: 5 x 2 + 5 y 2 = 125 5 x 2 + 5 y 2 = 125

If we expand the equation from Example 11.8 , ( x + 2 ) 2 + ( y − 1 ) 2 = 9 , ( x + 2 ) 2 + ( y − 1 ) 2 = 9 , the equation of the circle looks very different.

This form of the equation is called the general form of the equation of the circle .

General Form of the Equation of a Circle

The general form of the equation of a circle is

If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y . Then we can graph the circle using its center and radius.

Example 11.10

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 4 x − 6 y + 4 = 0 . x 2 + y 2 − 4 x − 6 y + 4 = 0 .

We need to rewrite this general form into standard form in order to find the center and radius.

Try It 11.19

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 6 x − 8 y + 9 = 0 . x 2 + y 2 − 6 x − 8 y + 9 = 0 .

Try It 11.20

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 + 6 x − 2 y + 1 = 0 . x 2 + y 2 + 6 x − 2 y + 1 = 0 .

In the next example, there is a y -term and a y 2 y 2 -term. But notice that there is no x -term, only an x 2 x 2 -term. We have seen this before and know that it means h is 0. We will need to complete the square for the y terms, but not for the x terms.

Example 11.11

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 + 8 y = 0 . x 2 + y 2 + 8 y = 0 .

Try It 11.21

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 2 x − 3 = 0 . x 2 + y 2 − 2 x − 3 = 0 .

Try It 11.22

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 12 y + 11 = 0 . x 2 + y 2 − 12 y + 11 = 0 .

Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

Distance-Midpoint Formulas and Circles
Finding the Distance and Midpoint Between Two Points
Completing the Square to Write Equation in Standard Form of a Circle

Section 11.1 Exercises

Practice makes perfect.

In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

( 2 , 0 ) ( 2 , 0 ) and ( 5 , 4 ) ( 5 , 4 )

( −4 , −3 ) ( −4 , −3 ) and ( 2 , 5 ) ( 2 , 5 )

( −4 , −3 ) ( −4 , −3 ) and ( 8 , 2 ) ( 8 , 2 )

( −7 , −3 ) ( −7 , −3 ) and ( 8 , 5 ) ( 8 , 5 )

( −1 , 4 ) ( −1 , 4 ) and ( 2 , 0 ) ( 2 , 0 )

( −1 , 3 ) ( −1 , 3 ) and ( 5 , −5 ) ( 5 , −5 )

( 1 , −4 ) ( 1 , −4 ) and ( 6 , 8 ) ( 6 , 8 )

( −8 , −2 ) ( −8 , −2 ) and ( 7 , 6 ) ( 7 , 6 )

( −3 , −5 ) ( −3 , −5 ) and ( 0 , 1 ) ( 0 , 1 )

( −1 , −2 ) ( −1 , −2 ) and ( −3 , 4 ) ( −3 , 4 )

( 3 , −1 ) ( 3 , −1 ) and ( 1 , 7 ) ( 1 , 7 )

( −4 , −5 ) ( −4 , −5 ) and ( 7 , 4 ) ( 7 , 4 )

In the following exercises, ⓐ find the midpoint of the line segments whose endpoints are given and ⓑ plot the endpoints and the midpoint on a rectangular coordinate system.

( 0 , −5 ) ( 0 , −5 ) and ( 4 , −3 ) ( 4 , −3 )

( −2 , −6 ) ( −2 , −6 ) and ( 6 , −2 ) ( 6 , −2 )

( 3 , −1 ) ( 3 , −1 ) and ( 4 , −2 ) ( 4 , −2 )

( −3 , −3 ) ( −3 , −3 ) and ( 6 , −1 ) ( 6 , −1 )

In the following exercises, write the standard form of the equation of the circle with the given radius and center ( 0 , 0 ) . ( 0 , 0 ) .

Radius: 2 2

Radius: 5 5

In the following exercises, write the standard form of the equation of the circle with the given radius and center

Radius: 1, center: ( 3 , 5 ) ( 3 , 5 )

Radius: 10, center: ( −2 , 6 ) ( −2 , 6 )

Radius: 2.5 , 2.5 , center: ( 1.5 , −3.5 ) ( 1.5 , −3.5 )

Radius: 1.5 , 1.5 , center: ( −5.5 , −6.5 ) ( −5.5 , −6.5 )

For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.

Center ( 3 , −2 ) ( 3 , −2 ) with point ( 3 , 6 ) ( 3 , 6 )

Center ( 6 , −6 ) ( 6 , −6 ) with point ( 2 , −3 ) ( 2 , −3 )

Center ( 4 , 4 ) ( 4 , 4 ) with point ( 2 , 2 ) ( 2 , 2 )

Center ( −5 , 6 ) ( −5 , 6 ) with point ( −2 , 3 ) ( −2 , 3 )

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

( x + 5 ) 2 + ( y + 3 ) 2 = 1 ( x + 5 ) 2 + ( y + 3 ) 2 = 1

( x − 2 ) 2 + ( y − 3 ) 2 = 9 ( x − 2 ) 2 + ( y − 3 ) 2 = 9

( x − 4 ) 2 + ( y + 2 ) 2 = 16 ( x − 4 ) 2 + ( y + 2 ) 2 = 16

( x + 2 ) 2 + ( y − 5 ) 2 = 4 ( x + 2 ) 2 + ( y − 5 ) 2 = 4

x 2 + ( y + 2 ) 2 = 25 x 2 + ( y + 2 ) 2 = 25

( x − 1 ) 2 + y 2 = 36 ( x − 1 ) 2 + y 2 = 36

( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25 ( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25

( x − 1 ) 2 + ( y − 3 ) 2 = 9 4 ( x − 1 ) 2 + ( y − 3 ) 2 = 9 4

x 2 + y 2 = 64 x 2 + y 2 = 64

x 2 + y 2 = 49 x 2 + y 2 = 49

2 x 2 + 2 y 2 = 8 2 x 2 + 2 y 2 = 8

6 x 2 + 6 y 2 = 216 6 x 2 + 6 y 2 = 216

In the following exercises, ⓐ identify the center and radius and ⓑ graph.

x 2 + y 2 + 2 x + 6 y + 9 = 0 x 2 + y 2 + 2 x + 6 y + 9 = 0

x 2 + y 2 − 6 x − 8 y = 0 x 2 + y 2 − 6 x − 8 y = 0

x 2 + y 2 − 4 x + 10 y − 7 = 0 x 2 + y 2 − 4 x + 10 y − 7 = 0

x 2 + y 2 + 12 x − 14 y + 21 = 0 x 2 + y 2 + 12 x − 14 y + 21 = 0

x 2 + y 2 + 6 y + 5 = 0 x 2 + y 2 + 6 y + 5 = 0

x 2 + y 2 − 10 y = 0 x 2 + y 2 − 10 y = 0

x 2 + y 2 + 4 x = 0 x 2 + y 2 + 4 x = 0

x 2 + y 2 − 14 x + 13 = 0 x 2 + y 2 − 14 x + 13 = 0

Writing Exercises

Explain the relationship between the distance formula and the equation of a circle.

Is a circle a function? Explain why or why not.

In your own words, state the definition of a circle.

In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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Authors: Lynn Marecek, Andrea Honeycutt Mathis
Publisher/website: OpenStax
Book title: Intermediate Algebra 2e
Publication date: May 6, 2020
Location: Houston, Texas
Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/11-1-distance-and-midpoint-formulas-circles

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11.2: Distance and Midpoint Formulas and Circles

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Learning Objectives

By the end of this section, you will be able to:

Use the Distance Formula

Use the midpoint formula.

Write the equation of a circle in standard form
Graph a circle

Before you get started, take this readiness quiz.

Find the length of the hypotenuse of a right triangle whose legs are $12$ and $16$ inches. If you missed this problem, review Example 2.34.
Factor: $x^{2}-18 x+81$. If you missed this problem, review Example 6.24.
Solve by completing the square: $x^{2}-12 x-12=0$. If you missed this problem, review Example 9.22.

$This figure shows two cones placed point to point. They are labeled nappes.$

There are four conics—the circle , parabola , ellipse , and hyperbola . The next figure shows how the plane intersecting the double cone results in each curve.

$Each of these four figures shows a double cone intersected by a plane. In the first figure, the plane is perpendicular to the axis of the cones and intersects the bottom cone to form a circle. In the second figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it intersects the base as well. Thus, the curve formed by the intersection is open at both ends. This is labeled parabola. In the third figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it does not intersect the base of the cone. Thus, the curve formed by the intersection is a closed loop, labeled ellipse. In the fourth figure, the plane is parallel to the axis, intersecting both cones. This is labeled hyperbola.$

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Example $\PageIndex{1}$

Use the rectangular coordinate system to find the distance between the points $(6,4)$ and $(2,1)$.

Exercise $\PageIndex{1}$

Use the rectangular coordinate system to find the distance between the points $(6,1)$ and $(2,-2)$.

Exercise $\PageIndex{2}$

Use the rectangular coordinate system to find the distance between the points $(5,3)$ and $(-3,-3)$.

$Figure shows a graph with a right triangle. The hypotenuse connects two points, (2, 1) and (6, 4). These are respectively labeled (x1, y1) and (x2, y2). The rise is y2 minus y1, which is 4 minus 1 equals 3. The run is x2 minus x1, which is 6 minus 2 equals 4.$

The method we used in the last example leads us to the formula to find the distance between the two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$.

When we found the length of the horizontal leg we subtracted $6−2$ which is $x_{2}-x_{1}$.

When we found the length of the vertical leg we subtracted $4−1$ which is $y_{2}-y_{1}$.

If the triangle had been in a different position, we may have subtracted $x_{1}-x_{2}$ or $y_{1}-y_{2}$. The expressions $x_{2}-x_{1}$ and $x_{1}-x_{2}$ vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say $\left|x_{2}-x_{1}\right|$ and $\left|y_{2}-y_{1}\right|$.

In the Pythagorean Theorem, we substitute the general expressions $\left|x_{2}-x_{1}\right|$ and $\left|y_{2}-y_{1}\right|$ rather than the numbers.

$\begin{array}{l c}{} & {a^{2}+b^{2}=c^{2}} \\ {\text {Substitute in the values. }}&{(|x_{2}-x_{1}|)^{2}+(|y_{2}-y_{1}|)^{2}=d^{2}} \\ {\text{Squaring the expressions makes}}&{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}=d^{2}} \\ \text{them positive, so we eliminate} \\\text{the absolute value bars.}\\ {\text{Use the Square Root Property.}}&{d=\pm\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}\\ {\text{Distance is positive, so eliminate}}&{d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}\\\text{the negative value.}\end{array}$

This is the Distance Formula we use to find the distance $d$ between the two points $(x_{1},y_{1})$ and $(x_{2}, y_{2})$.

Definition $\PageIndex{1}$

Distance Formula

The distance $d$ between the two points $(x_{1},y_{1})$ and $(x_{2}, y_{2})$ is

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Example $\PageIndex{2}$

Use the Distance Formula to find the distance between the points $(-5,-3)$ and $(7,2)$.

Write the Distance Formula.

Label the points, $\left( \begin{array}{c}{x_{1}, y_{1}} \\ {-5,-3}\end{array}\right)$, $\left( \begin{array}{l}{x_{2}, y_{2}} \\ {7,2}\end{array}\right)$ and substitute.

$d=\sqrt{(7-(-5))^{2}+(2-(-3))^{2}}$

$d=\sqrt{12^{2}+5^{2}}$ $d=\sqrt{144+25}$ $d=\sqrt{169}$ $d=13$

Exercise $\PageIndex{3}$

Use the Distance Formula to find the distance between the points $(-4,-5)$ and $(5,7)$.

Exercise $\PageIndex{4}$

Use the Distance Formula to find the distance between the points $(-2,-5)$ and $(-14,-10)$.

Example $\PageIndex{3}$

Use the Distance Formula to find the distance between the points $(10,−4)$ and $(−1,5)$. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Label the points, $\left( \begin{array}{c}{x_{1}, y_{1}} \\ {10,-4}\end{array}\right)$, $\left( \begin{array}{c}{x_{2}, y_{2}} \\ {-1,5}\end{array}\right)$ and substitute.

$d=\sqrt{(-1-10)^{2}+(5-(-4))^{2}}$

$d=\sqrt{(-11)^{2}+9^{2}}$ $d=\sqrt{121+81}$ $d=\sqrt{202}$

Since $202$ is not a perfect square, we can leave the answer in exact form or find a decimal approximation.

$d=\sqrt{202}$ or $d \approx 14.2$

Exercise $\PageIndex{5}$

Use the Distance Formula to find the distance between the points $(−4,−5)$ and $(3,4)$. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

$d=\sqrt{130}, d \approx 11.4$

Exercise $\PageIndex{6}$

Use the Distance Formula to find the distance between the points $(−2,−5)$ and $(−3,−4)$. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

$d=\sqrt{2}, d \approx 1.4$

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the $x$-coordinates and the average of the $y$-coordinates of the endpoints.

Definition $\PageIndex{2}$

Midpoint Formula

The midpoint of the line segment whose endpoints are the two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is

$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

To find the midpoint of a line segment, we find the average of the $x$-coordinates and the average of the $y$-coordinates of the endpoints.

Example $\PageIndex{4}$

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $(−5,−4)$ and $(7,2)$. Plot the endpoints and the midpoint on a rectangular coordinate system.

Exercise $\PageIndex{7}$

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $(−3,−5)$ and $(5,7)$. Plot the endpoints and the midpoint on a rectangular coordinate system.

$This graph shows a line segment with endpoints (negative 3, negative 5) and (5, 7) and midpoint (1, negative 1).$

Exercise $\PageIndex{8}$

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $(−2,−5)$ and $(6,−1)$. Plot the endpoints and the midpoint on a rectangular coordinate system.

$This graph shows a line segment with endpoints (negative 2, negative 5) and (6, negative 1) and midpoint (2, negative 3).$

Both the Distance Formula and the Midpoint Formula depend on two points, $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$. It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, is may be easier to remember the formulas.

$The distance formula is d equals square root of open parentheses x2 minus x1 close parentheses squared plus open parentheses y2 minus y1 close parentheses squared end of root. This is labeled subtract the coordinates. The midpoint formula is open parentheses open parentheses x1 plus x2 close parentheses upon 2 comma open parentheses y1 plus y2 close parentheses upon 2 close parentheses. This is labeled add the coordinates.$

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

$This figure shows a double cone and an intersecting plane, which form a circle.$

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, $(h,k)$, and the fixed distance is called the radius , $r$, of the circle.

Definition $\PageIndex{3}$

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center , $(h,k)$, and the fixed distance is called the radius , $r$, of the circle.

This is the standard form of the equation of a circle with center, $(h,k)$, and radius, $r$.

Definition $\PageIndex{4}$

The standard form of the equation of a circle with center, $(h,k)$, and radius, $r$, is

$Figure shows circle with center at (h, k) and a radius of r. A point on the circle is labeled x, y. The formula is open parentheses x minus h close parentheses squared plus open parentheses y minus k close parentheses squared equals r squared.$

Example $\PageIndex{5}$

Write the standard form of the equation of the circle with radius $3$ and center $(0,0)$.

Exercise $\PageIndex{9}$

Write the standard form of the equation of the circle with a radius of $6$ and center $(0,0)$.

$x^{2}+y^{2}=36$

Exercise $\PageIndex{10}$

Write the standard form of the equation of the circle with a radius of $8$ and center $(0,0)$.

$x^{2}+y^{2}=64$

In the last example, the center was $(0,0)$. Notice what happened to the equation. Whenever the center is $(0,0)$, the standard form becomes $x^{2}+y^{2}=r^{2}$.

Example $\PageIndex{6}$

Write the standard form of the equation of the circle with radius $2$ and center $(−1,3)$.

Exercise $\PageIndex{11}$

Write the standard form of the equation of the circle with a radius of $7$ and center $(2,−4)$.

$(x-2)^{2}+(y+4)^{2}=49$

Exercise $\PageIndex{12}$

Write the standard form of the equation of the circle with a radius of $9$ and center $(−3,−5)$.

$(x+3)^{2}+(y+5)^{2}=81$

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Example $\PageIndex{7}$

Write the standard form of the equation of the circle with center $(2,4)$ that also contains the point $(−2,1)$.

$This graph shows circle with center at (2, 4, radius 5 and a point on the circle minus 2, 1.$

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center $(2,4)$ and point $(−2,1)$

Use the Distance Formula to find the radius.

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Substitute the values. $\left( \begin{array}{l}{x_{1}, y_{1}} \\ {2,4}\end{array}\right), \left( \begin{array}{c}{x_{2}, y_{2}} \\ {-2,1}\end{array}\right)$

$r=\sqrt{(-2-2)^{2}+(1-4)^{2}}$

$r=\sqrt{(-4)^{2}+(-3)^{2}}$ $r=\sqrt{16+9}$ $r=\sqrt{25}$ $r=5$

Now that we know the radius, $r=5$, and the center, $(2,4)$, we can use the standard form of the equation of a circle to find the equation.

Use the standard form of the equation of a circle.

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Substitute in the values.

$(x-2)^{2}+(y-4)^{2}=5^{2}$

$(x-2)^{2}+(y-4)^{2}=25$

Exercise $\PageIndex{13}$

Write the standard form of the equation of the circle with center $(2,1)$ that also contains the point $(−2,−2)$.

$(x-2)^{2}+(y-1)^{2}=25$

Exercise $\PageIndex{14}$

Write the standard form of the equation of the circle with center $(7,1)$ that also contains the point $(−1,−5)$.

$(x-7)^{2}+(y-1)^{2}=100$

Graph a Circle

Any equation of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$ is the standard form of the equation of a circle with center, $(h,k)$, and radius, $r$ . We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from $x$ and $y$. In the next example, the equation has $x+2$, so we need to rewrite the addition as subtraction of a negative.

Example $\PageIndex{8}$

Find the center and radius, then graph the circle: $(x+2)^{2}+(y-1)^{2}=9$.

Exercise $\PageIndex{15}$

Find the center and radius, then
Graph the circle: $(x-3)^{2}+(y+4)^{2}=4$.
The circle is centered at $(3,-4)$ with a radius of $2$.

$This graph shows a circle with center at (3, negative 4) and a radius of 2.$

Exercise $\PageIndex{16}$

Graph the circle: $(x-3)^{2}+(y-1)^{2}=16$.
The circle is centered at $(3,1)$ with a radius of $4$.

$This graph shows circle with center at (3, 1) and a radius of 4.$

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of $x^{2}, y^{2}$ to be one.

Example $\PageIndex{9}$

Find the center and radius and then graph the circle, $4 x^{2}+4 y^{2}=64$.

Exercise $\PageIndex{17}$

Graph the circle: $3 x^{2}+3 y^{2}=27$
The circle is centered at $(0,0)$ with a radius of $3$.

$This graph shows circle with center at (0, 0) and a radius of 3.$

Exercise $\PageIndex{18}$

Graph the circle: $5 x^{2}+5 y^{2}=125$
The circle is centered at $(0,0)$ with a radius of $5$.

$This graph shows circle with center at (0, 0) and a radius of 5.$

If we expand the equation from Example 11.1.8, $(x+2)^{2}+(y-1)^{2}=9$, the equation of the circle looks very different.

$(x+2)^{2}+(y-1)^{2}=9$

Square the binomials.

$x^{2}+4 x+4+y^{2}-2 y+1=9$

Arrange the terms in descending degree order, and get zero on the right

$x^{2}+y^{2}+4 x-2 y-4=0$

This form of the equation is called the general form of the equation of the circle .

Definition $\PageIndex{5}$

The general form of the equation of a circle is

$x^{2}+y^{2}+a x+b y+c=0$

If we are given an equation in general form, we can change it to standard form by completing the squares in both $x$ and $y$. Then we can graph the circle using its center and radius.

Example $\PageIndex{10}$

Graph the circle: $x^{2}+y^{2}-4 x-6 y+4=0$

We need to rewrite this general form into standard form in order to find the center and radius.

Exercise $\PageIndex{19}$

Graph the circle: $x^{2}+y^{2}-6 x-8 y+9=0$.
The circle is centered at $(3,4)$ with a radius of $4$.

$This graph shows circle with center at (3, 4) and a radius of 4.$

Exercise $\PageIndex{20}$

Graph the circle: $x^{2}+y^{2}+6 x-2 y+1=0$
The circle is centered at $(-3,1)$ with a radius of $3$.

$This graph shows circle with center at (negative 3, 1) and a radius of 3.$

In the next example, there is a $y$-term and a $y^{2}$-term. But notice that there is no $x$-term, only an $x^{2}$-term. We have seen this before and know that it means $h$ is $0$. We will need to complete the square for the $y$ terms, but not for the $x$ terms.

Example $\PageIndex{11}$

Graph the circle: $x^{2}+y^{2}+8 y=0$

Exercise $\PageIndex{21}$

Graph the circle: $x^{2}+y^{2}-2 x-3=0$.
The circle is centered at $(-1,0)$ with a radius of $2$.

$This graph shows circle with center at (1, 0) and a radius of 2.$

Exercise $\PageIndex{22}$

Graph the circle: $x^{2}+y^{2}-12 y+11=0$.
The circle is centered at $(0,6)$ with a radius of $5$.

$This graph shows circle with center at (0, 6) and a radius of 5.$

Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

Distance-Midpoint Formulas and Circles
Finding the Distance and Midpoint Between Two Points
Completing the Square to Write Equation in Standard Form of a Circle

Key Concepts

Circle: A circle is all points in a plane that are a fixed distance from a fixed point in the plane. The given point is called the center, $(h,k)$, and the fixed distance is called the radius, $r$, of the circle.
Standard Form of the Equation a Circle: The standard form of the equation of a circle with center, $(h,k)$, and radius, $r$ , is

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Averages and range

How to find midpoint

How to find the midpoint

Here you will learn about how to find the midpoint, including finding the midpoint of a line segment using Cartesian coordinates, and finding a missing endpoint when the midpoint and other endpoint is given.

Students will first learn about how to find the midpoint as a part of geometry in 8 th grade.

What is the midpoint of a line?

The midpoint of a line segment is a point that lies exactly halfway between two points. It is the same distance from each endpoint of the straight line segment.

Sometimes we can work this out by inspection – this is easier with positive integer numbers.

For example, if we graph the two given points (2,2) and (8,6) on a coordinate plane, the midpoint is exactly halfway between the two, and lies at (5, 4).

It is easy to see that 5 is halfway between 2 and 8, and 4 is halfway between 2 and 6. You can easily visualize this using the graph below.

Midpoint of a line formula

If it is not easy to spot the midpoint, or the coordinates involve fractions or negative numbers, we can use the midpoint formula.

If the points \mathrm{A}\left(x_{1}, y_{1}\right) and \mathrm{B}\left(x_{2}, y_{2}\right) are the endpoints of a line segment, then the midpoint of the line segment joining the points A and B is \left(\cfrac{x_{1}+x_{2}}{2}, \cfrac{y_{1}+y_{2}}{2}\right) .

This looks complicated when written algebraically, but you’re basically calculating the (mean) average of both the x values and the y values.

Add the two x coordinates and divide by 2 to find the x coordinate of the midpoint, and add the two y coordinates and divide by 2 to find the y coordinate of the midpoint.

For example, given two points A (-1,2) and B (2,4), the midpoint (M) is exactly halfway between the two, and lies at (0.5, 3).

To calculate the midpoint,

The mean average of the x coordinates is \cfrac{-1+2}{2}=\cfrac{1}{2}=0.5

The mean average of the y coordinates is \cfrac{2+4}{2}=\cfrac{6}{2}=3

We can also apply the Pythagorean Theorem to find the distance between two given points. To do this, we form a right-angled triangle with the line segment as the hypotenuse.

Pythagoras’ theorem tells us that h^{2}=2^{2}+3^{2}, and therefore the length of the hypotenuse is \sqrt{2^{2}+3^{2}}=\sqrt{13}.

If you study coordinate geometry further in high school, you may come across the general distance formula:

d=\sqrt{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)} where (x_{1},y_{1}) and (x_{2},y_{2}) are the coordinates of the two points and d is the distance between them.

Finding a missing endpoint

You may not always be given both endpoints of the line. Sometimes you may be given one endpoint and the midpoint, and have to work out the other endpoint.

To get from the first endpoint (1,3) to the midpoint (3,7), move 2 in the x- direction and 4 in the y- direction. Repeat this again from the midpoint to find the coordinate of the other endpoint, which in this case would be (5,11).

Common Core State Standards

How does this relate to 8 th grade math?

Grade 8: Geometry (8.G.B.8) Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.

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How to find the midpoint of a line segment

In order to find the midpoint of the line segment joining the endpoints \text{A} and \text{B:}

Find the average of the \textbf{x} coordinates of the two endpoints.

Find the average of the \textbf{y} coordinates of the two endpoints.

Write down the coordinates of the point.

How to find the midpoint examples

Example 1: two positive integer endpoints.

Find the midpoint of the line segment joining the points (0,6) and (4, 10).

2 Find the average of the \textbf{y} coordinates of the two endpoints.

3 Write down the coordinates of the point.

In this case, it is quite easy to see the midpoint by inspection, particularly if working on a graph.

Example 2: two positive integer endpoints with a fractional answer

Find the midpoint of the line segment joining the points (1,5) and (6, 0).

It is generally OK to give coordinate pairs as short terminating decimals. Any longer or recurring decimals should be stated as fractions where possible, remembering to simplify your answer.

Example 3: coordinate pairs containing negative numbers

Find the midpoint of the line segment joining the points (-2,7) and (4, 10).

Graphically,

Example 4: coordinates containing decimals

Find the midpoint of the line segment joining the points (0.5, 3) and (4, 2.5).

How to find a missing endpoint

In order to find a missing endpoint when given one endpoint and the midpoint:

Work out how to get from the given endpoint to the midpoint.

Repeat this to get from the midpoint to the missing endpoint.

Write down the coordinates of the missing endpoint.

Example 5: finding a missing endpoint when given one endpoint and the midpoint

A line segment joins the points A and B, and has midpoint M.

Point A has coordinates (4, 8) and point M has coordinates (6, 9).

Find the coordinates of point B.

To get from A to M, we add 2 to the x- coordinate of A and add 1 to the y- coordinate of A.

To get from M to B, we add 2 to the x- coordinate of M and add 1 to the y- coordinate of M.

Therefore, the coordinates of point B are (8,10).

Example 6: finding a missing endpoint with negative coordinates

Point A has coordinates (-9, 4) and point M has coordinates (-6, -1).

To get from A to M, we add 3 to the x- coordinate of A and subtract 5 from the y- coordinate of A.

To get from M to B, we add 3 to the x- coordinate of M and subtract 5 from the y- coordinate of M.

Therefore, the coordinates of point B are (-3,-6).

Teaching tips for how to find the midpoint

Use visual aids to help illustrate the concept, such as number lines. This can provide foundational understanding for the concept of midpoints, before using a coordinate plane.
Provide students with step-by-step instructions on how to find the midpoint to refer back to when needed. These can be placed in a math journal or written on an anchor chart within the classroom.
Teach students to use interactive technology, like a midpoint calculator, to assist students with finding the midpoint using different avenues.

Easy mistakes to make

Finding the average of each point rather than the average of the \textbf{x} coordinates and average of the \textbf{y} coordinates For example, for the points (2, 3) and (5, 7), make sure you don’t do \cfrac{2+3}{2} and \cfrac{5+7}{2}.
Using the midpoint formula when given one endpoint and the midpoint If one endpoint is (3, 4) and the midpoint is (6, 2), make sure you work out how you get from the endpoint to the midpoint and repeat this, rather than using the midpoint formula.
Errors with negative number calculations If you’re not sure about your answer, draw a diagram and count the steps.

Related graphing linear equations lessons

Graphing linear equations
How to find the y- intercept
How to find the slope of a line
Distance formula
Linear interpolation

Practice how to find the midpoint questions

1) Find the midpoint of the line segment joining the points (2, 8) and (6, 12).

The average of the x coordinates is \cfrac{2+6}{2}=\cfrac{8}{2}=4 and the average of the y coordinates is \cfrac{8+12}{2}=\cfrac{20}{2}=10.

2) Find the midpoint of the line segment joining the points (4, 10) and (7, 5).

The average of the x coordinates is \cfrac{4+7}{2}=\cfrac{11}{2}=5.5 and the average of the y coordinates is \cfrac{10+5}{2}=\cfrac{15}{2}=7.5 .

3) Find the midpoint of the line segment joining the points (-2, 8) and (6, -2).

The average of the x coordinates is \cfrac{-2+6}{2}=\cfrac{4}{2}=2 and the average of the y coordinates is \cfrac{8+(-2)}{2}=\cfrac{6}{2}=3.

4) Find the midpoint of the line segment joining the points (3.5, 6) and (11, 8.5).

The average of the x coordinates is \cfrac{3.5+11}{2}=\cfrac{14.5}{2}=7.25 and the average of the y coordinates is \cfrac{6+8.5}{2}=\cfrac{14.5}{2}=7.5 .

5) A line segment joins the points A and B, and has midpoint M.

Point A has coordinates (4, 2) and point M has coordinates (9, 4)

To get from A to M, add 5 to the x coordinate and add 2 to the y coordinate.

Repeat this to get from M to B, so the coordinate of B is (14, 6).

6) A line segment joins the points A and B, and has midpoint M.

Point A has coordinates (3, 7) and point M has coordinates (-1, 10).

To get from point A to point M, subtract 4 from the x coordinate and add 3 to the y coordinate.

Repeat this to get from point M to point B, so the coordinate of point B is (-5, 13).

How to find the midpoint FAQs

A perpendicular bisector is a line or segment that crosses a given line segment precisely at its midpoint, creating a 90- degree angle (a right angle) with the original line. In more straightforward language, it is a line that divides a given line segment into two equal sections while meeting it at a right angle.

Yes, you can find the midpoint of both horizontal and vertical lines. The method for finding the midpoint depends on whether the line is horizontal or vertical. For a horizontal line segment, the x- coordinate of the two endpoints are different. For a vertical line segment, the y- coordinates of the two endpoints are different.

The next lessons are

Rate of change
Systems of equations
Number patterns

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High School Math : Midpoint and Distance Formulas

Study concepts, example questions & explanations for high school math, all high school math resources, example questions, example question #1 : how to find the midpoint of a line segment.