Exercise 3 Page 331

Let's add the midsegment, EF, to the diagram.

Note that O and C are on the x-axis which means they both have the y-coordinate. Therefore, the midpoint F must also have the y-coordinate 0. To calculate F's x-coordinate, we add the x-coordinates of O and C and divide by 2. x-coordinate F: 0+ 2p/2=pTherefore, F's coordinates are (p,0). To show that FE∥ OB, we have to prove that these segment have the same slope. m_(EF)=m_(OB) We already know the coordinates of O and B which means we can calculate the slope of OB. To determine the slope of EF, we need to figure out the coordinates of E. Since E is the midpoint of BC, we can find its coordinates by using the Midpoint Formula.

M(x_1+x_2/2,y_1+y_2/2)

SubstitutePoints

Substitute ( 2q,2r) & ( 2p,0)

M(2q+ 2p/2,2r+ 0/2)

▼

Simplify

FactorOut

Factor out 2

M(2(q+p)/2,2r+0/2)

AddTerms

Add terms

M(2(q+p)/2,2r/2)

SimpQuot

Simplify quotient

M(q+p,r)

Now we can calculate the slope of the segments using the Slope Formula.

Segment	Points	y_2-y_1/x_2-x_1	m
OB	( 2q,2r), ( 0,0)	2r- 0/2q- 0	r/q
EF	( q+p,r), ( p,0)	r- 0/( q+p)- p	r/q

Both segments have a slope of rq which means they are parallel. Finally, we can show that FE= 12OB by calculating the length of FE and OB using the Distance Formula.

Segment	Points	sqrt((x_2-x_1)^2+(y_2-y_1)^2)	d
OB	( 2q,2r), ( 0,0)	sqrt(( 2q- 0)^2+( 2r- 0)^2)	sqrt(4q^2+4r^2)
FE	( q+p,r), ( p,0)	sqrt((( q+p)- p)^2+( r- 0)^2)	sqrt(q^2+r^2)

By substituting the distances in the equation FE= 12OB, we can prove that the equation is true.

FE=1/2OB

SubstituteII

EF= sqrt(q^2+r^2), OB= sqrt(4q^2+4r^2)

sqrt(q^2+r^2)? =1/2( sqrt(4q^2+4r^2))

▼

Simplify right-hand side

FactorOut

Factor out 4

sqrt(q^2+r^2)? =1/2sqrt(4(q^2+r^2))

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

sqrt(q^2+r^2)? =1/2sqrt(4)sqrt(q^2+r^2)

CalcRoot

Calculate root

sqrt(q^2+r^2)? =1/2* 2(sqrt(q^2+r^2))

MoveRightFacToNumOne

1/b* a = a/b

sqrt(q^2+r^2)? =2/2sqrt(q^2+r^2)

QuotOne

a/a=1

sqrt(q^2+r^2)=sqrt(q^2+r^2)

As we can see, FE is half that of OB.