Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises - Page 457: 44

Answer

$(5,\pm\frac{16}{3})$

Work Step by Step

We plug in 5 for x, and then we solve for the corresponding y point that is that far away: $\sqrt{100 + y^2} = \mid{y}\mid +6 \\ 100 + y^2 = 36 + y^2 + 12\mid y \mid \\ 12 \mid y \mid = 64 \\ y = \pm \frac{16}{3}$ After finding these points, plot them on the graph. Once the points have been plotted on the graph, connect them using curvy lines (similar to those of parabolas).
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