Answer
False,
the distance is $4\sqrt{2}$.
Work Step by Step
$d(P, R)=\sqrt{(x_{2}-x_{\mathrm{I}})^{2}+(y_{2}-y_{1})^{2}}$
-----------
$P(0,0), R(4,4),$
$d(P, R)=\sqrt{(4-0)^{2}+(4-0)^{2}}$
$=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
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 192: 10
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
