Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 6 - Polygons and Quadrilaterals - 6-7 Polygons in the Coordinate Plane - Got It? - Page 401: 1

Answer

scalene

Work Step by Step

We use the distance formula to determine what type of triangle is pictured. The vertices of the triangle are $A(-2, 3)$, $B(2, 2)$, and $C(-2, -1)$. The distance formula is given by the following formula: $d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Let's determine the lengths of the different sides of the triangle. We'll look at $AB$ first: $AB = \sqrt {(2 - (-2))^2 + (2 - 3)^2}$ Simplify within the parentheses: $AB = \sqrt {(4)^2 + (-1)^2}$ Evaluate the exponents: $AB = \sqrt {16 + 1}$ Add what is underneath the radical: $AB = \sqrt {17}$ Let's look at the next side, $BC$: $BC = \sqrt {(-2 - 2)^2 + (-1 - 2))^2}$ Simplify within the parentheses: $BC = \sqrt {(-4)^2 + (-3)^2}$ Evaluate the exponents: $BC = \sqrt {16 + 9}$ Add what is underneath the radical: $BC = \sqrt {25}$ Evaluate to solve: $BC = 5$ Let's look at $CA$: $CA = \sqrt {(-2 - (-2))^2 + (-1 - 3)^2}$ Simplify within the parentheses: $CA = \sqrt {(0)^2 + (-4)^2}$ Evaluate the exponents: $CA = \sqrt {0 + 16}$ Add what is underneath the radical: $CA = \sqrt {16}$ Evaluate to solve: $CA = 4$ All three sides have different lengths; therefore, this triangle is scalene.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.