College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.1 - Rectangular Coordinates and Graphs - 2.1 Exercises - Page 178: 19


Yes, this is a right triangle.

Work Step by Step

The distance between A and B: $$\sqrt{(-6-0)^2+(-4-(-2))^2}$$ $$\sqrt{36+4}$$ $$\sqrt{40}$$ The distance between B and C: $$\sqrt{(0-(-10))^2+(-2-8)^2}$$ $$\sqrt{100+100}$$ $$\sqrt{200}$$ The distance between C and D: $$\sqrt{(-6-(-10))^2+(-4-8)^2}$$ $$\sqrt{16+144}$$ $$\sqrt{160}$$ Now, check that these three points form a right triangle using the Pythagorean Theorem. $$160+40=200$$ $$200=200$$ Yes, this is a right triangle.
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