## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 1, Equations and Graphs - Section 1.1 - The Coordinate Plane - 1.1 Exercises - Page 92: 41

#### Answer

$a.\quad$ Please see "step-by-step" $b.\quad$ Area = $10$

#### Work Step by Step

a. Reading the ccoordinates from the graph, we have $A=(2,2), B=(3, -1)$, and $C=(-3, -3)$. Calculate the sides' lengths: $d(A, B)=\sqrt{(3-2)^{2}+(-1-2)^{2}}=\sqrt{1^{2}+(-3)^{2}}=\sqrt{1+9}=\sqrt{10}.$ $d(C, B)=\sqrt{(3-(-3))^{2}+(-1-(-3))^{2}}=\sqrt{6^{2}+2^{2}}=\sqrt{36+4}\\=\sqrt{40}=2\sqrt{10}.$ $d(A, C)=\sqrt{(-3-2)^{2}+(-3-2)^{2}}=\sqrt{(-5)^{2}+(-5)^{2}}\\=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$. Note that $[d(A, B)]^{2}+[d(C, B)]^{2}=10+4\cdot 10=50$ $[d(A, C)]^{2}=25\cdot 2=50,$ so, $[d(A, B)]^{2}+[d(C, B)]^{2}=[d(A, C)]^{2}$. By the converse of the Pythagorean Theorem, we conclude that the triangle is a right triangle. b. The area of the triangle is $\displaystyle \frac{1}{2}\cdot d(C, B)\cdot d(A, B)=\frac{1}{2}\cdot\sqrt{10}\cdot 2\sqrt{10}=10$.

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