Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises: 25a

Answer

Isosceles Triangle

Work Step by Step

A(0,0) B (4,0) and C(2,5) Using distance formula x1=0,y1=0,x2=4,y2=0 then, $\overline{AB}$ = $\sqrt (x2-x1)^{2} + (y2-y1)^{2}$ = $\sqrt (4-0)^{2} + (0-0)^{2}$ =4 units Similarly, $\overline{BC}$ = $\sqrt (2-4)^{2} + (5-0)^{2}$ =$\sqrt 2^{2} + 5^{2}$ = $\sqrt 4+25$ = $\sqrt 29$ $\overline{CA}$ = $\sqrt (2-0)^{2} + (5-0)^{2}$ =$\sqrt 2^{2} + 5^{2}$ = $\sqrt 4+25$ = $\sqrt 29$ Therefore $\overline{BC}$ = $\overline{CA}$ in triangle ABC therefore triangle ABC is isosceles.
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