Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX B - Coordinate Geometry and Lines - B Exercises: 5

Answer

$2\sqrt {37}$

Work Step by Step

To find the distance between $(2,5)$ and $(4,-7)$, we use the distance formula: $d=\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$. We plug in the given $x$ and $y$ values: $d=\sqrt {(4-2)^{2}+(-7-5)^{2}}$. We then simplify to arrive at the answer: $d=\sqrt {2^{2}+12^{2}}=\sqrt {148} = 2\sqrt {37}$.
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