College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.1 - The Coordinate Plane - 1.1 Exercises - Page 92: 42


The triangle is a right triangle. Area =$\displaystyle \frac{41}{2}$

Work Step by Step

$d(A, B) = \sqrt{(11-6)^{2}+(-3-(-7))^{2}} = \sqrt{5^{2}+4^{2}}$ $= \sqrt{25+16} = \sqrt{41}.$ $d(A, C) = \sqrt{(2-6)^{2}+(-2-(-7))^{2}} = \sqrt{(-4)^{2}+5^{2}}$ $= \sqrt{16+25} = \sqrt{41}$; $d(B, C)=\sqrt{(2-11)^{2}+(-2-(-3))^{2}}=\sqrt{(-9)^{2}+1^{2}}$ $=\sqrt{81+1}=\sqrt{82}$. $[d(A, B)]^{2}+[d(A, C)]^{2}=41+41=82,$ $[d(B, C)]^{2}=82$, Since$\qquad [d(A, B)]^{2}+[d(A, C)]^{2}=[d(B, C)]^{2}$, we conclude that the triangle is a right triangle. The Area is$ =\displaystyle \frac{1}{2}$ (base)$\times$(height). $A = \displaystyle \frac{1}{2}(\sqrt{41}\cdot\sqrt{41})=\frac{41}{2}$.
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