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Core Connections Geometry, 2013 View details

Chapter Closure

Finish the assignment

Exercise 116 Page 204

a To find the slope of the line through the points, we substitute them in the Slope Formula and simplify.

m = y_2 - y_1/x_2 - x_1

SubstitutePoints

Substitute ( 2,1) & ( - 2,7)

m = 1 - 7/2 - ( - 2)

▼

Simplify right-hand side

SubNeg

a-(- b)=a+b

m = 1-7/2+2

AddSubTerms

Add and subtract terms

m = - 6/4

ReduceFrac

a/b=.a /2./.b /2.

m = - 3/2

MoveNegDenomToFrac

Put minus sign in front of fraction

m = - 3/2

The slope is - 32.

b To find the distance between the points, we substitute the points in the Distance Formula and simplify.

d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)

SubstitutePoints

Substitute ( 2,1) & ( -2,7)

d = sqrt(( 2 - ( - 2))^2 + ( 1 - 7)^2)

▼

Simplify right-hand side

SubNeg

a-(- b)=a+b

d = sqrt((2 + 2)^2 + (1 - 7)^2)

AddSubTerms

Add and subtract terms

d = sqrt(4^2 + (- 6 )^2)

CalcPow

Calculate power

d = sqrt(16 + 36)

AddTerms

Add terms

d = sqrt(52)

SplitIntoFactors

Split into factors

d = sqrt(4* 13)

SqrtProd

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

d = sqrt(4)sqrt(13)

CalcRoot

Calculate root

d = 2sqrt(13)

UseCalc

Use a calculator

d = 7.21110...

RoundDec

Round to 1 decimal place(s)

d ≈ 7.2

The distance is about 7.2 units.

c From Part A, we know that the slope of the line is - 32. With this, we have half of what we need to write the equation of the line in slope-intercept form.

y=- 3/2x+bTo complete the equation, we also have to find the y-intercept, b. We can do that by substituting one of the points into the equation and solving for b.

y=- 3/2x+b

SubstituteII

x= 2, y= 1

1=- 3/2( 2)+b

▼

Solve for b

MoveRightFacToNum

a/c* b = a* b/c

1=- 6/2+b

CalcQuot

Calculate quotient

1=- 3+b

AddEqn

LHS+3=RHS+3

4=b

RearrangeEqn

Rearrange equation

b=4

Now we can complete the equation. y=- 3/2x+4

d To write the equation of a line perpendicular to RP and through point P, we first need to determine its slope. Two lines are perpendicular when their slopes are opposite reciprocals. This means that the product of a given slope and the slope of a line perpendicular to it will be -1.

m_1*m_2=-1 From Part A, we have determined the slope of RP. By substituting this slope into the equation, we can solve for the slope of the perpendicular line, m_2.

m_1 * m_2 = - 1

Substitute

m_1= - 3/2

- 3/2* m_2 = - 1

▼

Solve for m_2

MultEqn

LHS * (- 1)=RHS* (- 1)

3/2* m_2 = 1

MultEqn

LHS * 2=RHS* 2

3* m_2 = 2

DivEqn

.LHS /3.=.RHS /3.

m_2 = 2/3

Any line perpendicular to the given line will have a slope of 23 which means we can write the equation in the following form. y=2/3x+b To find b, we substitute the coordinates of P in this equation and solve for b.

y=2/3x+b

SubstituteII

x= 2, y= 1

1=2/3( 2)+b

▼

Solve for b

MoveRightFacToNum

a/c* b = a* b/c

1=4/3+b

MultEqn

LHS * 3=RHS* 3

3=4+3b