College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.1 - The Distance and Midpoint Formulas - 2.1 Assess Your Understanding - Page 155: 55

Answer

$\sqrt{17},\ 2\sqrt{5},\ \sqrt{29}$

Work Step by Step

Midpoint of A and B: $M_{AB}=\displaystyle \left(\frac{0+6}{2},\frac{0+0}{2}\right)=(3,0)$ $d(C,M_{AB})=\sqrt{(0-4)^{2}+(3-4)^{2}}=\sqrt{(-4)^{2}+(-1)^{2}}=\sqrt{16+1}=\sqrt{17}$ $M_{AC}=\displaystyle \left(\frac{0+4}{2},\frac{0+4}{2}\right)=(2,2)$ $d(B,M_{AC})=\sqrt{(2-6)^{2}+(2-0)^{2}}=\sqrt{(-4)^{2}+2^{2}}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$ $M_{BC}=\displaystyle \left(\frac{6+4}{2},\frac{0+4}{2}\right)=(5,2)$ $d(A,M_{BC})=\sqrt{(2-0)^{2}+(5-0)^{2}}=\sqrt{2^{2}+5^{2}}=\sqrt{4+25}=\sqrt{29}$
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