Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 3 - Linear and Quadratic Functions - Chapter Review - Cumulative Review - Page 165: 1

Answer

$d=5\sqrt2.$ $M=\left(\frac{3}{2},\frac{1}{2}\right)$.

Work Step by Step

The distance formula from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. The midpoint $M$ of the line segment from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is: $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$. Hence: $d=\sqrt{(4-(-1))^2+(-2-3)^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt2.$ $M=\left(\frac{-1+4}{2},\frac{3+(-2)}{2}\right)=\left(\frac{3}{2},\frac{1}{2}\right)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.