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: 33

Answer

See graph. $BC^2=AB^2+AC^2$, $10\ unit^2$.

Work Step by Step

Step 1. See graph. Step 2. $AB=\sqrt {(4-0)^2+(-3+3)^2}=4$, $BC=\sqrt {(0-4)^2+(-3-2)^2}=\sqrt {41}$, $AC=\sqrt {(4-4)^2+(-3-2)^2}=5$, Step 3. Check $BC^2=AB^2+AC^2$, thus it is a right triangle, Step 4. $Area=\frac{1}{2}(4)(5)=10\ unit^2$.
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