PRECALCULUS:GRAPHICAL,...-NASTA ED.
PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780134672090
Author: Demana
Publisher: PEARSON
Expert Solution & Answer
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Chapter 6.1, Problem 62E
Solution

a.

To prove:

  OM1=12OA+12OB OM2=12OC+12OB OM3=12OA+12OC .

Given information:

The medians of a triangle are in the ratio 1:2 .

  M1,M2 and M3 are the midpoints of the sides of the triangle.

Proof:

The given triangle ABC ,

  PRECALCULUS:GRAPHICAL,...-NASTA ED., Chapter 6.1, Problem 62E , additional homework tip  1

From the exercise 61b the triangle OAB , M1 is the midpoint of AB

  x=12y=1x=112=12

Substituting the values,

  OM1=xOA+yOB=12OA+12OB

Similarly,

From triangle OBC ,

  OM2=xOB+yOC=12OB+12OC

From triangle OCA ,

  OM3=xOC+yOA=12OC+12OA

Hence OM1=12OA+12OB ,  OM2=12OC+12OB and OM3=12OA+12OC has been proved.

b.

To prove:

  2OM1+OC=OA+OB+OC2OM2+OA=OA+OB+OC2OM3+OB=OA+OB+OC

Given information:

The medians of a triangle are in the ratio 1:2 .

  M1,M2 and M3 are the midpoints of the sides of the triangle.

Proof:

The given triangle ABC ,

  PRECALCULUS:GRAPHICAL,...-NASTA ED., Chapter 6.1, Problem 62E , additional homework tip  2

From part (a), it is known that

  OM1=12OA+12OB

Multiply by 2 on both sides to get,

  2OM1=OA+OB

Adding OC on both sides to get,

  2OM1+OC=OA+OB+OC

Similarly, need to find the other two identities

  OM2=12OB+12OC2OM2=OC+OB2OM2+OA=OC+OB+OAOM3=12OA+12OC2OM3=OA+OC2OM3+OC=OA+OC+OB

Therefore, the identities 2OM1+OC=OA+OB+OC , 2OM2+OA=OA+OB+OC and 2OM3+OB=OA+OB+OC has been proved.

c.

To prove: Why part (b) establishes the desired outcome.

Given information:

The medians of a triangle are in the ratio 1:2 .

  M1,M2 and M3 are the midpoints of the sides of the triangle.

Proof:

The given triangle ABC ,

  PRECALCULUS:GRAPHICAL,...-NASTA ED., Chapter 6.1, Problem 62E , additional homework tip  3

Now, draw the two medians

  PRECALCULUS:GRAPHICAL,...-NASTA ED., Chapter 6.1, Problem 62E , additional homework tip  4

From part (b), it is known that

  2OM1+OC=2OM3+OB=2OM2+OA=OA+OB+OC

Considering parallelogram BOCD ,

Let’s take the vector on AD ,

  OA+OB+OC=OA+OD

Let’s take the vector on CE ,

  OC+OA+OB=OC+OE

Since it is situated on two different directions, then

  OA+OB+OC=0

From part (b),

  2OM1+OC=OA+OB+OC2OM1+OC=0OC=2OM1

Similarly,

  OB=2OM3OA=2OM2

Thus, the vector OB and OM3 passes through the same direction at point O .

Therefore, the third median passes through the intersection of the other because the points B, O, M3 are collinear.

Chapter 6 Solutions

PRECALCULUS:GRAPHICAL,...-NASTA ED.

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