/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 (a) plot the points, (b) find th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 34

(a) plot the points, (b) find the distance between the points, and (c) find the midpoint of the line segment joining the points. $$(2,10),(10,2)$$

Short Answer

Expert verified
The distance between the points (2,10) and (10,2) is approximately 11.31 units. The midpoint of the line segment joining these points is (6,6).

Step by step solution

01

Plotting the points on a coordinate plane

The points provided are (2,10) and (10,2). Plot these points on a graph. The first point (2,10) is located at 2 units on the x-axis and 10 units on the y-axis. Similarly, the other point (10,2) is located at 10 units on the x-axis and 2 units on the y-axis.
02

Calculating the distance between the points

To find the distance between the points, use the distance formula which is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). Substituting the coordinates of the given points, the formula becomes \(\sqrt{(10-2)^2+(2-10)^2}\) which simplifies to \(\sqrt{8^2+(-8)^2}\), and further simplifies to \(\sqrt{64+64}\) which equals \(\sqrt{128}\). The exact distance is \(\sqrt{128}\), but it can be approximated as 11.31.
03

Finding the midpoint of the line segment

To find the midpoint, use the midpoint formula which is \(\[(x_1+x_2)/2,(y_1+y_2)/2\]\). Substituting the coordinates of the given points, the formula becomes \(\[(2+10)/2,(10+2)/2\]\) which simplifies to [6,6]. So, the midpoint of the line segment joining the points (2,10) and (10,2) is (6,6).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine whether the statement is true or false. Justify your answer. It is possible for an odd function to have the interval \([0, \infty)\) as its domain.

Sketch the graph of the function. $$k(x)=\left\\{\begin{array}{ll}2 x+1, & x \leq-1 \\\2 x^{2}-1, & -11\end{array}\right.$$

Consider \(f(x)=\sqrt{x-1}\) and \(g(x)=\frac{1}{\sqrt{x-1}}\) Why are the domains of \(f\) and \(g\) different?

(a) Write the linear function \(f\) such that it has the indicated function values and (b) Sketch the graph of the function. $$f(-5)=-1, \quad f(5)=-1$$

Use a graphing utility to graph the function. Be sure to choose an appropriate viewing window. $$f(x)=\frac{5}{6}-\frac{2}{3} x$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.