The shortest distance from point K to line is the length of a segment perpendicular to from point K. Draw a perpendicular segment from K to.

Similar documents
Use the graph shown to determine whether each system is consistent or inconsistent and if it is independent or dependent.

2-5 Rational Functions

This is a function because no vertical line can be drawn so that it intersects the graph more than once.

HFCC Math Lab Intermediate Algebra 1 SLOPE INTERCEPT AND POINT-SLOPE FORMS OF THE LINE

1-7 Inverse Relations and Functions

3.5 Day 1 Warm Up. Graph each line. 3.4 Proofs with Perpendicular Lines

Mid-Chapter Quiz: Lessons 1-1 through 1-4

Math 154 Elementary Algebra. Equations of Lines 4.4

Mid-Chapter Quiz: Lessons 3-1 through 3-4. Solve each system of equations. SOLUTION: Add both the equations and solve for x.

3-6 Lines in the Coordinate Plane

Did You Find a Parking Space?

slope rise run Definition of Slope

2-1 Power and Radical Functions

Geometry Unit 2: Linear. Section Page and Problems Date Assigned

Geometry Pre AP Graphing Linear Equations

Practice Test (page 391) 1. For each line, count squares on the grid to determine the rise and the run. Use slope = rise

Section Graphs and Lines

.(3, 2) Co-ordinate Geometry Co-ordinates. Every point has two co-ordinates. Plot the following points on the plane. A (4, 1) D (2, 5) G (6, 3)

Algebra 1 Semester 2 Final Review

List of Topics for Analytic Geometry Unit Test

Mathematics (

Point-Slope Form of an Equation

Tangent line problems

Writing Equations of Parallel and Perpendicular Lines

Mid-Chapter Quiz: Lessons 2-1 through 2-3

Chapter 1 Section 1 Solving Linear Equations in One Variable

2-5 Graphing Special Functions. Graph each function. Identify the domain and range. SOLUTION:

Chapter 3 Practice Test

Practice Test - Chapter 1

SLOPE A MEASURE OF STEEPNESS through 7.1.5

This is a function because no vertical line can be drawn so that it intersects the graph more than once.

SNAP Centre Workshop. Graphing Lines

Section 1.5. Finding Linear Equations

Practice Test - Chapter 6

of Straight Lines 1. The straight line with gradient 3 which passes through the point,2

Math-2. Lesson 3-1. Equations of Lines

Hot X: Algebra Exposed

Intensive Math-Algebra I Mini Lesson MA.912.A.3.10

Geometry Unit 5 Geometric and Algebraic Connections. Table of Contents

Section 1.1 The Distance and Midpoint Formulas

If you place one vertical and cross at the 0 point, then the intersection forms a coordinate system. So, the statement is true.

Exploring Slope. We use the letter m to represent slope. It is the ratio of the rise to the run.

Linear Functions. College Algebra

Derivatives. Day 8 - Tangents and Linearizations

Section 3.1 Objective 1: Plot Points in the Rectangular Coordinate System Video Length 12:35

The Rectangular Coordinate System and Equations of Lines. College Algebra

2-4 Writing Linear Equations. Write an equation in slope-intercept form for the line described. 9. slope passes through (0, 5) SOLUTION:

Vertical Line Test a relationship is a function, if NO vertical line intersects the graph more than once

3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS

AICE Mathematics AS Level Summer Review

Writing Linear Functions

UNIT 4 NOTES. 4-1 and 4-2 Coordinate Plane

Chapter 1. Linear Equations and Straight Lines. 2 of 71. Copyright 2014, 2010, 2007 Pearson Education, Inc.

JUST THE MATHS SLIDES NUMBER 5.2. GEOMETRY 2 (The straight line) A.J.Hobson

10-2 Circles. Warm Up Lesson Presentation Lesson Quiz. Holt Algebra2 2

Math 1313 Prerequisites/Test 1 Review

Math 2 Coordinate Geometry Part 3 Inequalities & Quadratics

Test Name: Chapter 3 Review

2-1 Power and Radical Functions

Summer Math Assignments for Students Entering Algebra II

WRITING AND GRAPHING LINEAR EQUATIONS ON A FLAT SURFACE #1313

2-5 Postulates and Paragraph Proofs

Unit 6: Connecting Algebra and Geometry Through Coordinates

Summer Math Assignments for Students Entering Integrated Math

Math 3 Coordinate Geometry part 1 Unit November 3, 2016

Review for Mastery Using Graphs and Tables to Solve Linear Systems

Also available in Hardcopy (.pdf): Coordinate Geometry

CHAPTER 2 REVIEW COORDINATE GEOMETRY MATH Warm-Up: See Solved Homework questions. 2.2 Cartesian coordinate system

Equations of Lines - 3.4

Chapter P: Preparation for Calculus

10-7 Special Segments in a Circle. Find x. Assume that segments that appear to be tangent are tangent. 1. SOLUTION: 2. SOLUTION: 3.

AP Calculus AB Summer Review Packet

SLOPE A MEASURE OF STEEPNESS through 2.1.4

2-7 Parent Functions and Transformations. Identify the type of function represented by each graph. ANSWER: linear ANSWER: absolute value

3-8 Solving Systems of Equations Using Inverse Matrices. Determine whether each pair of matrices are inverses of each other. 13.

5.1 Introduction to the Graphs of Polynomials

11-9 Areas of Circles and Sectors. CONSTRUCTION Find the area of each circle. Round to the nearest tenth. 1. Refer to the figure on page 800.

Math 2 Coordinate Geometry Part 1 Slope & Transformations

Algebra Unit 2: Linear Functions Notes. Slope Notes. 4 Types of Slope. Slope from a Formula

1-2 Analyzing Graphs of Functions and Relations

Study Guide and Review

You should be able to plot points on the coordinate axis. You should know that the the midpoint of the line segment joining (x, y 1 1

Name: NOTES 5: LINEAR EQUATIONS AND THEIR GRAPHS. Date: Period: Mrs. Nguyen s Initial: LESSON 5.1 RATE OF CHANGE AND SLOPE. A. Finding rates of change

1.5 Equations of Lines and Planes in 3-D

5-3 Polynomial Functions

Classwork/Homework. Midterm Review. 1) 9 more than the product of a number and 12 2) 5 less than a number squared is twelve.

graphing_9.1.notebook March 15, 2019

Intro. To Graphing Linear Equations

5. In the Cartesian plane, a line runs through the points (5, 6) and (-2, -2). What is the slope of the line?

Section 2.2 Graphs of Linear Functions

Graphs and Linear Functions

UNIT NUMBER 5.2. GEOMETRY 2 (The straight line) A.J.Hobson

Lakeview Christian Academy Summer Math Packet For Students Entering Algebra 2

Practice Test - Chapter 4. Classify each triangle as acute, equiangular, obtuse, or right.

Graphing Linear Equations

SOLUTION: Because the fractions have a common denominator, compare the numerators. 5 < 3

2. Write the point-slope form of the equation of the line passing through the point ( 2, 4) with a slope of 3. (1 point)

Name. Center axis. Introduction to Conic Sections

Coordinate Geometry. Coordinate geometry is the study of the relationships between points on the Cartesian plane

Lesson 3 Practice Problems

Transcription:

8. Find the distance between each pair of parallel lines with the given equations. Copy each figure. Construct the segment that represents the distance indicated. 12. K to The shortest distance from point K to line is the length of a segment perpendicular to from point K. Draw a perpendicular segment from K to. The two lines have the coefficient of x, zero. So, the slopes are zero. Therefore, the lines are horizontal lines passing through y = 7 and y = 3 respectively. The line perpendicular will be vertical. Thus, the distance, is the difference in the y-intercepts of the two lines. Then perpendicular distance between the two horizontal lines is 7 ( 3) = 10 units. 10 units esolutions Manual - Powered by Cognero Page 1

14. APPLY MATH Rondell is crossing the courtyard in front of his school. Three possible paths are shown in the diagram. Which of the three paths shown is the shortest? Explain your reasoning. COORDINATE GEOMETRY Find the distance from P to 17. Line contains points ( 2, 1) and (4, 1). Point P has coordinates (5, 7). Use the slope formula to find the slope of the line.let (x 1 ) = ( 2, 1)and (x 2, y 2 ) = (4, 1). The shortest possible distance would be the perpendicular distance from one side of the courtyard to the other. Since Path B is the closest to 90, it is the shortest of the three paths shown. Path B; The shortest possible distance would be the perpendicular distance from one side of the courtyard to the other. Since Path B is the closest to 90, it is the shortest of the three paths shown. Use the slope and any one of the points to write the equation of the line. Let (x 1 ) = ( 2, 1). The slope of an equation perpendicular to will be undefined, and hence the line will be a vertical line. The equation of a vertical line through (5, 7) is x = 5. The point of intersection of the two lines is (5, 1). between the points (5, 1) and (5, 7). Let (x 1 ) = (5, 1) and (x 2, y 2 ) = (5, 7). Therefore, the distance between the line and the point is 6 units. 6 units esolutions Manual - Powered by Cognero Page 2

Find the distance between each pair of parallel lines with the given equations. 22. 21. The two lines are horizontal lines and for each equation, the coefficient of x-term is zero. So, the slopes are zero. Therefore, the line perpendicular to the parallel lines will be vertical. The distance of the vertical line between the parallel lines, will be the difference in the y-intercepts. To find the perpendicular distance between the two horizontal lines subtract 2 from 4 to get 4 ( 2) = 6 units. 6 units 23. The two lines are vertical and of the form x = a. So, the slopes are undefined. Therefore, the lines are vertical lines passing through x = 3 and x = 7 respectively. The line perpendicular to each line will be horizontal. The distance will be the difference in the x-intercepts. To find the perpendicular distance between the two horizontal lines subtract 3 from 7 to get 7 3 = 4 units 4 units To find the distance between the parallel lines, we need to find the length of the perpendicular segment between the two parallel lines. Pick a point on one of the equation, and write the equation of a line perpendicular through that point. Then use this perpendicular line and other equation to find the point of intersection. Find the distance between the two point using the distance formula. Step 1: Find the equation of the line perpendicular to each of the lines. The slope of a line perpendicular to both the lines will be Consider the y-intercept of any of the two lines and write the equation of the perpendicular line through it. The y-intercept of the line y = 5x + 4 is (0, esolutions Manual - Powered by Cognero Page 3

4). So, the equation of a line with slope and a y- intercept of 4 is 24. Step 2: Find the intersections of the perpendicular line and each of the other lines. To find the point of intersection of the perpendicular and the second line, solve the two equations. The left sides of the equations are the same. So, equate the right sides and solve for x. To find the distance between the parallel lines, we need to find the length of the perpendicular segment between the two parallel lines. Pick a point on one of the equation, and write the equation of a line perpendicular through that point. Then use this perpendicular line and other equation to find the point of intersection. Find the distance between the two point using the distance formula. Step 1: Find the equation of the line perpendicular to each of the lines. Use the value of x to find the value of y. So, the point of intersection is (5, 3). Step 3: Find the length of the perpendicular between points between the points (5, 3) and (0, 4). Let (x 1 ) = (5, 3) and (x 2, y 2 ) = (0, 4). The slope of a line perpendicular to both the lines will be 3. Consider the y-intercept of any of the two lines and write the equation of the perpendicular line through it. The y-intercept of the line is (0, 2). So, the equation of a line with slope 3 and a y-intercept of 2 is Step 2: Find the intersections of the perpendicular line and each of the other lines. To find the point of intersection of the perpendicular and the second line, solve the two equations. The left sides of the equations are the same. So, equate the right sides and solve for x. Therefore, the distance between the two lines is Use the value of x to find the value of y. esolutions Manual - Powered by Cognero Page 4

So, the point of intersection is Step 3: Find the length of the perpendicular between points. between the points and (0, 2). Let (x 1, y 1 ) = and (x 2, y 2 ) = (0, 2). Therefore, the distance between the two lines is esolutions Manual - Powered by Cognero Page 5