Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 103: 50

Answer

$$BM=AM=CM=\frac{\sqrt{a^2+b^2}}{2}$$

Work Step by Step

Let's calculate the distances using $a$ and $b$. But first we will need $M$ coordinates (We can calculate it using midpoint method of $AB$). $M=(\frac{a+0}{2}, \frac{0+b}{2})=(\frac{a}{2}, \frac{b}{2})$ $BM=\sqrt{(\frac{a}{2}-0)^2+(\frac{b}{2}-b)^2}=\sqrt{(\frac{a}{2})^2+(-\frac{b}{2})^2}=\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}=\frac{\sqrt{a^2+b^2}}{2}$ $AM=\sqrt{(\frac{a}{2}-a)^2+(\frac{b}{2}-0)^2}=\sqrt{(-\frac{a}{2})^2+(\frac{b}{2})^2}=\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}=\frac{\sqrt{a^2+b^2}}{2}$ $CM=\sqrt{(\frac{a}{2}-0)^2+(\frac{b}{2}-0)^2}=\sqrt{(\frac{a}{2})^2+(\frac{b}{2})^2}=\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}=\frac{\sqrt{a^2+b^2}}{2}$ As we can see above $BM=AM=CM=\frac{\sqrt{a^2+b^2}}{2}$
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