   # The problem of finding the perpendicular bisector of a line segment presents itself often in the study of analytic geometry. As with any problem of writing the equation of a line, you must determine the slope of the line and a point that the line passes through. A perpendicular bisector passes through the midpoint of the line segment and has a slope that is the negative reciprocal of the slope of the line segment The problem can be solved as follows: Find the perpendicular bisector of the line segment between the points ( 1 , − 2 ) and ( 7 , 8 ) . The midpoint of the line segment is ( 1 + 7 2 , − 2 + 8 2 ) = ( 4 , 3 ) . The slope of line segment is m = 8 − ( − 2 ) 7 − 1 = 10 6 = 5 3 . Hence the perpendicular bisector will pass through the point ( 4 , 3 ) and the slope of m = − 3 5 . y − 3 = − 3 5 ( x − 4 ) 5 ( y − 3 ) = − 3 ( x − 4 ) 5 y − 15 = − 3 x + 12 3 x + 5 y = 27 Thus the equation of perpendicular bisector of the line segment between the points ( 1 , − 2 ) and ( 7 , 8 ) is 3 x + 5 y = 27 . Find the perpendicular bisector of the line segment between the points for the following. Write the equation in standard form. (a) ( − 1 , 2 ) and ( 3 , 0 ) (b) ( 6 , − 10 ) and ( − 4 , 2 ) (c) ( − 7 , − 3 ) and ( 5 , 9 ) (d) ( 0 , 4 ) and ( 12 , − 4 ) ### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
Publisher: Cengage Learning
ISBN: 9781285195728

#### Solutions

Chapter
Section ### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
Publisher: Cengage Learning
ISBN: 9781285195728
Chapter 7.4, Problem 85PS
Textbook Problem
1 views

## The problem of finding the perpendicular bisector of a line segment presents itself often in the study of analytic geometry. As with any problem of writing the equation of a line, you must determine the slope of the line and a point that the line passes through. A perpendicular bisector passes through the midpoint of the line segment and has a slope that is the negative reciprocal of the slope of the line segment The problem can be solved as follows:Find the perpendicular bisector of the line segment between the points ( 1 , − 2 ) and ( 7 , 8 ) .The midpoint of the line segment is ( 1 + 7 2 , − 2 + 8 2 ) = ( 4 , 3 ) .The slope of line segment is m = 8 − ( − 2 ) 7 − 1 = 10 6 = 5 3 .Hence the perpendicular bisector will pass through the point ( 4 , 3 ) and the slope of m = − 3 5 . y − 3 = − 3 5 ( x − 4 ) 5 ( y − 3 ) = − 3 ( x − 4 ) 5 y − 15 = − 3 x + 12 3 x + 5 y = 27 Thus the equation of perpendicular bisector of the line segment between the points ( 1 , − 2 ) and ( 7 , 8 ) is 3 x + 5 y = 27 .Find the perpendicular bisector of the line segment between the points for the following. Write the equation in standard form.(a) ( − 1 , 2 ) and ( 3 , 0 ) (b) ( 6 , − 10 ) and ( − 4 , 2 ) (c) ( − 7 , − 3 ) and ( 5 , 9 ) (d) ( 0 , 4 ) and ( 12 , − 4 )

To determine

(a)

To find:

The equation of perpendicular bisector of the line segment between the given points.

(1,2) and (3,0)

### Explanation of Solution

Approach:

Standard form of a linear equation is given by,

Ax+By=C

Where B and C are integers, and A is a non-negative integer.

Formula used:

The mid-point of the line segment is given by,

(x1+x22,y1+y22)

Where, (x1,y1) and (x2,y2) are points on the line.

Slope of line is given by,

m=y2y1x2x1

Where, (x1,y1) and (x2,y2) are points on the line.

Point-Slope Form of an equation is,

yy1=m(xx1)

Where, (x1,y1) and m are points and slope of the line respectively.

Calculation:

The given points are,

(1,2) and (3,0)

Substitute 1 for x1, 3 for x2, 2 for y1 and 7 for y2 in the above mentioned mid-point formula.

To determine

(b)

To find:

The equation of perpendicular bisector of the line segment between the given points.

(6,10) and (4,2)

To determine

(c)

To find:

The equation of perpendicular bisector of the line segment between the given points.

(7,3) and (5,9)

To determine

(d)

To find:

The equation of perpendicular bisector of the line segment between the given points.

(0,4) and (12,4)

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
Differentiate the function. f(x)=240

Calculus (MindTap Course List)

Calculate y'. 20. y = exsec x

Single Variable Calculus: Early Transcendentals, Volume I

In Exercises 89-94, determine whether the statement is true or false. If it is true, explain why it is true. If...

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

Evaluate each expression: 12918648+7

Elementary Technical Mathematics

Evaluate 6 8 24 30

Study Guide for Stewart's Multivariable Calculus, 8th

Elimination of the parameter in x = 2t3/2, y = t2/3 gives: x4 = 16y9 16x4 = y4 x3 = 8y4 8x3 = y4

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

Define a statistic and a parameter and explain the role that each plays in inferential statistics.

Research Methods for the Behavioral Sciences (MindTap Course List)

Reminder Round all answers to two decimal places unless otherwise indicated. WendysAccording to a report in The...

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

(a) Construct a linear first-order differential equation of the form xy + 3y = g(x) for which y = x3 + c/x3 is ...

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

The data from exercise 3 follow. What is the value of the standard error of the estimate? Test for a significa...

Modern Business Statistics with Microsoft Office Excel (with XLSTAT Education Edition Printed Access Card) (MindTap Course List) 