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 7: 30

Answer

See graph. $AC^2=AB^2+BC^2$, $100\ unit^2$.

Work Step by Step

Step 1. See graph. Step 2. $AB=\sqrt {(-2-12)^2+(5-3)^2}=\sqrt {200}$, $BC=\sqrt {(12-10)^2+(3+11)^2}=\sqrt {200}$, $AC=\sqrt {(-2-10)^2+(5+11)^2}=\sqrt {400}=20$, Step 3. Check $AC^2=AB^2+BC^2$, thus it is a right triangle, Step 4. $Area=\frac{1}{2}(\sqrt {200})(\sqrt {200})=100\ unit^2$.
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