Answer
$(-11,-3)$ and $(13,-3)$.
Work Step by Step
Let the point(s) be $(x,-3)$, its distance to $(1,2)$ is $d=\sqrt {(x-1)^2+(-3-2)^2}=13$ or $(x-1)^2=144$, thus $x=1\pm12=-11,13$ which give answers $(-11,-3)$ and $(13,-3)$.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter F - Foundations: A Prelude to Functions - Section F.1 The Distance and Midpoint Formulas - F.1 Assess Your Understanding - Page 7: 46
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