College Algebra (10th Edition)

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

Chapter 2 - Review Exercises - Page 196: 29

Answer

$AB = BC = \sqrt{13}$, So triangle ABC is isosceles

Work Step by Step

Given three points A = (3, 4), B = (1, 1), and C = (- 2, 3) To show they are vertices of isosceles triangle Finding lengths AB, BC, AC $AB = \sqrt{(3-1)^2 + (4-1)^2} = \sqrt{13}$ $AC = \sqrt{(3-(-2))^2 + (4-3)^2} = \sqrt{26}$ $BC = \sqrt{(1-(-2))^2 +(1-3)^2} = \sqrt{13}$ Since $AB = BC = \sqrt{13}$, Triangle ABC is isosceles
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