unit 1 geometry basics homework 2 distance & midpoint formulas

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The distance and midpoint formulas

• Midpoint II
• Midpoint III

The distance formula is used to find the distance between two points in the coordinate plane. We'll explain this using an example below

We want to calculate the distance between the two points (-2, 1) and (4, 3). We could see the line drawn between these two points is the hypotenuse of a right triangle. The legs of this triangle would be parallel to the axes which mean that we can measure the length of the legs easily.

We'll get the length of the distance d by using the Pythagorean Theorem

$$d^{2}=2^{2}+6^{2}=4+36=40$$

$$d=\sqrt{40}\approx 6.32$$

This method can be used to determine the distance between any two points in a coordinate plane and is summarized in the distance formula

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

The point that is at the same distance from two points A (x 1 , y 1 ) and B (x 2 , y 2 ) on a line is called the midpoint. You calculate the midpoint using the midpoint formula

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

We can use the example above to illustrate this

$$ m =\left ( \frac{4+(-2)}{2} \right ),\: \: \left ( \frac{3+1}{2} \right )=$$

$$=\left ( \frac{2}{2} \right ),\: \: \left ( \frac{4}{2} \right )=\begin{pmatrix} 1,\: 2 \end{pmatrix}$$

Video lesson

Calculate the distance between the two points

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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 segment 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 segment 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 segment 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 segment 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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Geometry Basics (Geometry Curriculum - Unit 1) | All Things Algebra®

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This Geometry Basics Unit Bundle contains guided notes, homework assignments, three quizzes, dictionary, study guide and a unit test that cover the following topics:

• Points, Lines, and Planes

• Segment Addition Postulate

• The Distance Formula

• The Midpoint Formula

• Partitioning a Segment

• Naming and Classifying Angles

• Angle Addition Postulate

• Angle Relationships (Adjacent, Vertical, Complementary, Supplementary, Linear Pair)

• Solving for Missing Measures using Algebra

• Special Relationships: Perpendicular and Angle Bisectors

• Constructions (Perpendicular bisectors, perpendicular line through a point on the line, perpendicular line through a point not on the line, line parallel to a given line through a given point, angle bisector, congruent angles)

Note: This unit was updated on 8/25/19 to include partitioning a segment. If you do not teach this topic, I included the old unit without this topic in the download as well.

ADDITIONAL COMPONENTS INCLUDED:

(1) Links to Instructional Videos: Links to videos of each lesson in the unit are included. Videos were created by fellow teachers for their students using the guided notes and shared in March 2020 when schools closed with no notice.  Please watch through first before sharing with your students. Many teachers still use these in emergency substitute situations. (2) Editable Assessments: Editable versions of each quiz and the unit test are included. PowerPoint is required to edit these files. Individual problems can be changed to create multiple versions of the assessment. The layout of the assessment itself is not editable. If your Equation Editor is incompatible with mine (I use MathType), simply delete my equation and insert your own.

(3) Google Slides Version of the PDF: The second page of the Video links document contains a link to a Google Slides version of the PDF. Each page is set to the background in Google Slides. There are no text boxes;  this is the PDF in Google Slides.  I am unable to do text boxes at this time but hope this saves you a step if you wish to use it in Slides instead! 

This resource is included in the following bundle(s):

Geometry First Semester Notes Bundle

Geometry Curriculum

Geometry Curriculum (with Activities)

More Geometry Units:

Unit 2 – Logic and Proof

Unit 3 – Parallel and Perpendicular Lines

Unit 4 – Congruent Triangles

Unit 5 – Relationships in Triangles

Unit 6 – Similar Triangles

Unit 7 – Right Triangles and Trigonometry Unit 8 – Polygons and Quadrilaterals

Unit 9 – Transformations

Unit 10 – Circles

Unit 11 – Volume and Surface Area Unit 12 – Probability

LICENSING TERMS: This purchase includes a license for one teacher only for personal use in their classroom. Licenses are non-transferable , meaning they can not be passed from one teacher to another. No part of this resource is to be shared with colleagues or used by an entire grade level, school, or district without purchasing the proper number of licenses. If you are a coach, principal, or district interested in transferable licenses to accommodate yearly staff changes, please contact me for a quote at [email protected].

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Unit 1 Geometry Basics Homework 2 Distance And Midpoint Formulas

Don't miss, which of the following is not commonly studied in psychology, what does competition mean in biology, what is three dimensional geometry, what do letters mean in math, why are there different branches of chemistry, how to support ell students in math, what are these called in math, what are the types of bonds in chemistry, applying the distance and midpoint formulas.

The distance and midpoint formulas can help us find the distance and midpoint between two endpoints on a coordinate plane. The midpoint of a line segment formula between two coordinate pairs and

The distance of that same line segment denoted by coordinate pairs and

With a little more practice, you can navigate these formulas with ease.

Presentation On Theme: Lesson 811 Distance And Midpoint Formula Presentation Transcript:

1 Lesson 8.11 distance and midpoint formula Lesson 1-3: Formulas

2 Homework 1. P and R 2. S and T Use the Distance Formula to find the distance of:1. P and R2. S and TUse the Midpoint formula to find the midpoint between:3. W and X4. B and DLesson 1-3: Formulas

3 The Distance FormulaThe distance d between any two points with coordinates and is given by the formula d =Lesson 1-3: Formulas

4 Example 1: Finding Distances in the Coordinate Plane Use the Distance Formula to find the distance, to the nearest tenth, fromD to E.

5 Example 1 ContinuedUse the Distance Formula. Substitute thevalues for the coordinates of D and E into theDistance Formula.

6 Example 2: Using the Distance Formula Find the length for FG and JK.Then determine whether FG JK.Step 1 Find the coordinates of each point.F, G, J, K

7 Example 2 ContinuedStep 2 Use the Distance Formula.

8 Check It Out! Example 3Find EF and GH. Then determine if EF GH.Step 1 Find the coordinates of each point.E, F, G, H

9 Check It Out! Example 3 Continued Step 2 Use the Distance Formula.

10 Example 4: Sports Application A player throws the ball from first base to a point located between third base and home plate and 10 feet from third base.What is the distance of the throw, to the nearest tenth?

13 Example 1: Finding the Coordinates of a Midpoint Find the coordinates of the midpoint of PQ with endpoints P and Q.=

14 Check It Out! Example 1Find the coordinates of the midpoint of EF with endpoints E and F.

Unit 1 Geometry Basics Homework 3 Distance And Midpoint Formulas Answer Key

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Directions: Find the distance between each pair of points.1. 1-4.6) and 3. and

Directions: Find the coordinates of the midpoint of the segment given its endpoints.6. /15, 8) and B7. M and N[-2.7)8. P and Q13.-5)Gina 2014-2017

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How To Solve The Midpoint Formula

Youll use the midpoint formula to find the middle point between any two points. To do this, the formula looks at the coordinates of the endpoints and then finds the average value between the x-coordinates and the y-coordinates.

In the graph below, we have a line segment between the two coordinate points. is , and

The midpoint of the line segment is expressed as:

The red dot in the middle represents the new coordinate pair for the midpoint:

Lets substitute the coordinate values from above into the equation:

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The Distance And Midpoint Formulas

The distance formula is used to find the distance between two points in the coordinate plane. We’ll explain this using an example below

We want to calculate the distance between the two points and . We could see the line drawn between these two points is the hypotenuse of a right triangle. The legs of this triangle would be parallel to the axes which mean that we can measure the length of the legs easily.

We’ll get the length of the distance d by using the Pythagorean Theorem

$$d^=2^+6^=4+36=40$$

$$d=\sqrt\approx 6.32$$

This method can be used to determine the distance between any two points in a coordinate plane and is summarized in the distance formula

$$d=\sqrt-x_ \right )^+\left ^}$$

The point that is at the same distance from two points A and B on a line is called the midpoint. You calculate the midpoint using the midpoint formula

$$m =\left ,\: \: \left $$

We can use the example above to illustrate this

$$ m =\left ,\: \: \left =$$

$$=\left ,\: \: \left =\begin 1,\: 2 \end$$

How To Use The Distance And Midpoint Formulas On A Coordinate Plane

With the distance and midpoint formulas, you can find the distance and midpoint between any two points on a coordinate plane.

The distance formula gives you the distance d, expressed as a single value, between the two endpoints:

The midpoint formula gives you the midpoint, expressed as an ordered pair , between the two endpoints:

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How To Solve The Distance Formula

The distance formula works for any two given points. Let’s take a look at this example with coordinate points and :

If we imagine the two end points as two vertices on a triangle, we can see how the distance formula works. Remember that the hypotenuse of a right triangle , when squared, equals the sum of the square of the two legs. This is where the distance formula comes from.

Heres where the triangle is in our diagram:

In this example, the green and blue dotted sides of the triangle are perpendicular lines that form a right triangle.

This blue side of the triangle is 4 units long, and the green side of the triangle is 2 units long, as we can see on the graph. But how long is the red side? Its the distance between our two original points.

Lets apply the Pythagorean theorem , c² = a² + b². Well substitute the red line for c , the hypotenuse, and the green and blue lines for sides a and b .

Well take the square root of each side:

Finally, we substitute the colors. Red becomes D, or the distance.

Now, look at our coordinate pairs. Well refer to as and as .

Green is the change in the x-values, so well subtract the x-values of the two coordinate pairs,.

Likewise, the value for blue, the change in y, is .

The distance formula emerges when we substitute these values:

The length of the hypotenuse here is the distance between our end points. Lets substitute the points to figure out the answer:

Then, we simplify according to the order of operations, PEMDAS :

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Unit 1 Geometry Basics Homework 2 Distance Midpoint Formula

Displaying top 8 worksheets found for - Unit 1 Geometry Basics Homework 2 Distance Midpoint Formula .

Some of the worksheets for this concept are Midpoint and distance formulas, Geometry basics, 3 the midpoint formula, Using midpoint and distance formulas, Unit 1 geometry basics geometry, Big ideas chapter 1 basics of geometry, Geometry, Geometry distance and midpoint word problems.

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