Chapter 39: Chapter 39

Problem 679

Write an equation of the locus of points equidistant from the points \(\mathrm{P}_{1}(2,2)\) and \(\mathrm{P}_{2}(6,2)\).

Short Answer

Expert verified

The equation of the locus of points equidistant from the points \(\mathrm{P}_{1}(2,2)\) and \(\mathrm{P}_{2}(6,2)\) is \(x=4\).

See the step by step solution

Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the midpoint of the line segment joining \(\mathrm{P}_{1}\) with \(\mathrm{P}_{2}\)

The midpoint of a line segment with endpoints \((x_1,y_1)\) and \((x_2,y_2)\) is given by the formula \(M(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\). Applying this formula to the given points: \(M(\frac{2+6}{2},\frac{2+2}{2}) = M(4,2)\) So the midpoint of the line segment joining \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\) is \(M(4,2)\).

Step 2: Find the slope of the line segment joining \(\mathrm{P}_{1}\) with \(\mathrm{P}_{2}\)

The slope of a line segment with endpoints \((x_1,y_1)\) and \((x_2,y_2)\) can be calculated using the formula \(m = \frac{y_2-y_1}{x_2-x_1}\). Applying this formula to the given points: \(m = \frac{2-2}{6-2} = \frac{0}{4} = 0\) The slope of the line segment joining \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\) is \(0\).

Step 3: Determine the slope of the perpendicular bisector

The slope of the perpendicular bisector will be the negative reciprocal of the slope of the line segment joining the points \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\). Since the slope of the line segment is \(0\), the perpendicular bisector will have a slope that is undefined or a vertical line.

Step 4: Use the point-slope form to find the equation of the perpendicular bisector

Since the slope of the perpendicular bisector is undefined, this implies that the perpendicular bisector is a vertical line. A vertical line has a constant \(x\) value. Since the midpoint of the line segment joining \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\) is \(M(4,2)\), the equation of the perpendicular bisector will be: \(x=4\) So, the equation of the locus of points equidistant from the points \(\mathrm{P}_{1}(2,2)\) and \(\mathrm{P}_{2}(6,2)\) is \(x=4\).

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Recommended explanations on Math Textbooks

Select your language

Chapter 39: Chapter 39

Write an equation of the locus of points equidistant from the points \(\mathrm{P}_{1}(2,2)\) and \(\mathrm{P}_{2}(6,2)\).

Short Answer

Step by step solution

Unlock all solutions

Step 1: Find the midpoint of the line segment joining \(\mathrm{P}_{1}\) with \(\mathrm{P}_{2}\)

Step 2: Find the slope of the line segment joining \(\mathrm{P}_{1}\) with \(\mathrm{P}_{2}\)

Step 3: Determine the slope of the perpendicular bisector

Step 4: Use the point-slope form to find the equation of the perpendicular bisector

Access millions of textbook solutions in one place

Most popular questions from this chapter

More chapters from the book ‘Geometry’

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Chapter 17

Chapter 18

Chapter 19

Chapter 20

Chapter 21

Chapter 22

Chapter 23

Chapter 24

Chapter 25

Chapter 26

Chapter 27

Chapter 28

Chapter 29

Chapter 30

Chapter 31

Chapter 32

Chapter 33

Chapter 34

Chapter 35

Chapter 36

Chapter 37

Chapter 38

Chapter 40

Chapter 41

Chapter 42

Chapter 43

Chapter 44

Chapter 48

Chapter 49

Chapter 50

Chapter 51

Chapter 52

Join over 22 million students in learning with our Vaia App

Recommended explanations on Math Textbooks

Pure Maths

Probability and Statistics

Decision Maths

Calculus

Mechanics Maths

Statistics