Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.1 Rectangular Coordinates and Graphs - 2.1 Exercises - Page 194: 45

Answer

See explanations.

Work Step by Step

Step 1. Based on the given conditions, we have $x_m=\frac{x_1+x_2}{2}$ and $y_m=\frac{y_1+y_2}{2}$ Step 2. We can find: $x_m-x_1=\frac{-x_1+x_2}{2}$, $y_m-y_1=\frac{-y_1+y_2}{2}$, $x_m-x_2=\frac{x_1-x_2}{2}$, $y_m-y_2=\frac{y_1-y_2}{2}$, Step 3. Use the distance formula, we have $d(P,M)=\sqrt {(x_m-x_1)^+(y_m-y_1)^2}=\frac{1}{2}\sqrt {(x_1-x_2)^2+(y_1-y_2)^2}=\frac{1}{2}d(P,Q)$ Step 4. Use the distance formula, we have $d(M,Q)=\sqrt {(x_m-x_2)^+(y_m-y_2)^2}=\frac{1}{2}\sqrt {(x_1-x_2)^2+(y_1-y_2)^2}=\frac{1}{2}d(P,Q)$ Step 5. Thus we have $d(P,M)+d(M,Q)=d(P,Q)$ and $d(P,M)=d(M,Q)=\frac{1}{2}d(P,Q)$
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.