Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - 9.1 Apply the Distance and Midpoint Formulas - 9.1 Exercises - Skill Practice - Page 618: 46

Answer

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Work Step by Step

Let's note $A(x_1,y_1)$, $B(x_2,y_2)$. Determine the distance $d_{AM}$ and the distance $d_{BM}$ using the distance formula: $$\begin{align*} d_{AM}&=\sqrt{(x_A-x_M)^2+(y_A-y_M)^2}\\ &=\sqrt{\left(x_1-\dfrac{x_1+x_2}{2}\right)^2+\left(y_1-\dfrac{y_1+y_2}{2}\right)^2}\\ &=\sqrt{\dfrac{(x_1-x_2)^2}{4}+\dfrac{(y_1-y_2)^2}{4}}\\ &=\dfrac{1}{2}\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ &=\dfrac{1}{2}d_{AB}. \end{align*}$$ $$\begin{align*} d_{BM}&=\sqrt{(x_B-x_M)^2+(y_B-y_M)^2}\\ &=\sqrt{\left(x_2-\dfrac{x_1+x_2}{2}\right)^2+\left(y_2-\dfrac{y_1+y_2}{2}\right)^2}\\ &=\sqrt{\dfrac{(x_2-x_1)^2}{4}+\dfrac{(y_2-y_1)^2}{4}}\\ &=\dfrac{1}{2}\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ &=\dfrac{1}{2}d_{AB}. \end{align*}$$ Because $d_{AM}=d_{BM}$, we got that $M$ is equidistant from each endpoint $A$ and $B$. Because $d_{AM}=d_{BM}=\frac{1}{2}d_{AB}$, it means that $M$ lies on the segment $\overline{AB}$ and is its midpoint.
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