Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter F - Foundations: A Prelude to Functions - Section F.1 The Distance and Midpoint Formulas - F.1 Assess Your Understanding - Page 6: 18

Answer

$d(p_1,P_2)=\sqrt {10}$ units

Work Step by Step

$P_1=(-1,1)$ so $x_1 = -1 , y_1= 1$ $P_2=(2,2)$ so $x_2 = 2 , y_2= 2$ The distance between $P_1$ and $P_2$ can be found using the distance formula. $d=\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}$ $d=\sqrt {(2-(-1))^2+(2-1)^2}$ $d=\sqrt{3^2+1^2}$ $d=\sqrt {9+1}$ $d=\sqrt {10}$
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