Exercise 1

Chapter 9, Section 9-1, Page 385
enVision Geometry 1st Edition by Al Cuoco
ISBN: 9780328931583

Solution

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Step 1
1 of 2

First, we determine the midpoint of AC\overline{AC} which we denote by XX. Using A(1,6)A(1,6) and C(8,3)C(8,3),

X(1+82,6+32)X(92,92)X\left(\dfrac{1+8}{2},\dfrac{6+3}{2}\right)\to X\left(\dfrac{9}{2},\dfrac{9}{2}\right)

Next, we determine the midpoint of BC\overline{BC} which we denote by YY. Using B(3,1)B(3,1) and C(8,3)C(8,3),

Y(3+82,1+32)Y(112,2)Y\left(\dfrac{3+8}{2},\dfrac{1+3}{2}\right)\to Y\left(\dfrac{11}{2},2\right)

The length of the segment connecting the midpoints of AC\overline{AC} and BC\overline{BC} is XYXY. Using the Distance Formula,

XY=(11292)2+(292)2XY=\sqrt{\left(\dfrac{11}{2}-\dfrac{9}{2}\right)^2+\left(2-\dfrac{9}{2}\right)^2}

XY=(1)2+(52)2XY=\sqrt{\left(1\right)^2+\left(-\dfrac{5}{2}\right)^2}

XY=1+254XY=\sqrt{1+\dfrac{25}{4}}

XY=294XY=\sqrt{\dfrac{29}{4}}

XY=292\color{#c34632}{XY=\dfrac{\sqrt{29}}{2}}

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