College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.1 - The Distance and Midpoint Formulas - 2.1 Assess Your Understanding - Page 156: 56

Answer

$C=(2\sqrt{3},2)$ or $(-2\sqrt{3},2)$ (two such triangles are possible)

Work Step by Step

As seen in previous exercises, the median is a line segment connecting a vertex and the midpoint of the opposite side. The median of an equilateral triangle is also its altitude. The altitude is perpendicular to the side. The given side has length $4$ and is on the y-axis, so the median is parallel to the x-axis. This means the third point has the same y-coordinate as the midpoint of the side. $A=(0,0), B=(0,4)$ $M_{AB}=\displaystyle \left(\frac{0+0}{2},\frac{4+0}{2}\right)=(0,2).$ $C=(x,2)$ The x-coordinate of the third point will be one altitude to the left, or one altitude to the right. The altitude (height) of an equilateral triangle is $h=\displaystyle \frac{a\sqrt{3}}{2}=\frac{4\sqrt{3}}{2}=2\sqrt{3}.$ The x-coordinate can be $+2\sqrt{3}$ or $-2\sqrt{3}.$ $C=(2\sqrt{3},2)$ or $(-2\sqrt{3},2)$ (two such triangles are possible)
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