Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set - Page 282: 95


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Work Step by Step

a. Let $x_m=\frac{x_1+x_2}{2}$ and $y_m=\frac{y_1+y_2}{2}$, we have $(x_1-x_m)^2=(x_2-x_m)^2=(\frac{x_1-x_2}{2})^2$ and $(y_1-y_m)^2=(y_2-y_m)^2=(\frac{y_1-y_2}{2})^2$. We can find $\bar {AM}=\sqrt {(x_1-x_m)^2+(y_1-y_m)^2}=\sqrt {(\frac{x_1-x_2}{2})^2+(\frac{y_1-y_2}{2})^2}$ and $\bar {MB}=\sqrt {(x_2-x_m)^2+(y_2-y_m)^2}=\sqrt {(\frac{x_1-x_2}{2})^2+(\frac{y_1-y_2}{2})^2}$ b. We can calculate the three distances as: $\bar {AM}=\bar {MB}=\frac{1}{2}\sqrt {(x_1-x_2)^2+(y_1-y_2)^2}$ and $\bar {AB}=\sqrt {(x_1-x_2)^2+(y_1-y_2)^2}$. Since $\bar {AB}=\bar {AM}+\bar {MB}$, the three points A, B, M are collinear.
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