## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 1, Equations and Graphs - Section 1.1 - The Coordinate Plane - 1.1 Exercises - Page 92: 39

#### Answer

The sides $AC$ and $CB$ have the same length, so the triangle is isosceles.

#### Work Step by Step

Find the lengths of all three sides. $d(A, B)=\sqrt{(-3-0)^{2}+(-1-2)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$. Find the lengths of all three sides. $d(A, B)=\sqrt{(-3-0)^{2}+(-1-2)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}$ $=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$. $d(C, B)=\sqrt{(-3-(-4))^{2}+(-1-3)^{2}}=\sqrt{1^{2}+(-4)^{2}}$ $=\sqrt{1+16}=\sqrt{17}$. $d(A, C)=\sqrt{(0-(-4))^{2}+(2-3)^{2}}=\sqrt{4^{2}+(-1)^{2}}$ $=\sqrt{16+1}=\sqrt{17}$. The sides $AC$ and $CB$ have the same length, so the triangle is isosceles.

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