Exercise 34 Page 317

Let's start by graphing the triangle using the given coordinates.

To find the circumcenter, we need to find the intersection of the perpendicular bisectors of the sides of the triangle. This means that we need to know the equations of at least two of them. Recall that a bisector cuts something in half, so we want to find lines that are perpendicular to the sides at their midpoints.

Finding Perpendicular Bisectors

Since we can find perpendicular bisectors of any of two sides, let's do it for DE and DF. Let's first find their midpoints. To do so, we can use the Midpoint Formula.

Side	Points	M(x_1+x_2/2,y_1+y_2/2)	Midpoint
DE	( -9,-5), ( -5,-9)	G(-9+( -5)/2,-5+( -9)/2)	G(-7,-7)
DF	( -9,-5), ( -2,-2)	H(-9+( -2)/2,-5+( -2)/2)	H(-11/2,-7/2)

Let's add these midpoints to our graph.

Once we know the midpoints, we are ready to find the equations of perpendicular bisectors.

Perpendicular Bisector of DE

According to the Slopes of Perpendicular Lines Theorem, the product of the slopes of perpendicular lines is -1. Let's find the slope of DE first. Since we know the endpoints of DE, we can find its slope using the Slope Formula. m = y_2-y_1/x_2-x_1 Let's substitute the given endpoints D( -9, -5) and E( -5, -9).

m_(DE) = y_2-y_1/x_2-x_1

SubstitutePoints

Substitute ( -9,-5) & ( -5,-9)

m_(DE)=-9-( -5)/-5-( -9)

▼

Simplify right-hand side

SubNeg

a-(- b)=a+b

m_(DE)=-9+5/-5+9

AddTerms

Add terms

m_(DE)=-4/4

CalcQuot

Calculate quotient

m_(DE)=-1

We found that the slope of DE is -1. Now we can find the slope of the perpendicular bisector from the product. Let's call it m_(p_1). -1* m_(p_1) = -1 ⇔ m_(p_1) = 1 The slope of the perpendicular bisector of DE is 1. y=1x+b ⇒ y=x+b To complete its equation, we need the y-intercept. We can use the fact that the line passes through the midpoint G( -7, -7 ).

y=x+b

SubstituteII

x= -7, y= -7

-7= -7+b

▼

Solve for b

AddEqn

LHS+7=RHS+7

0=b

RearrangeEqn

Rearrange equation

b=0

Therefore, the equation of the perpendicular bisector of DE is y=x.

Perpendicular Bisector of DF

Similarly we can find the equation of the perpendicular bisector of DF. Let's first find the slope of DF using the Slope Formula.

m_(DF) = y_2-y_1/x_2-x_1

SubstitutePoints

Substitute ( -9,-5) & ( -2,-2)

m_(DF)=-2-( -5)/-2-( -9)

▼

Simplify right-hand side

SubNeg

a-(- b)=a+b

m_(DF) = -2+5/-2+9

AddTerms

Add terms

m_(DF) = 3/7

Once we know the slope of DF, we can find the slope of its perpendicular bisector from the product. Let's call it m_(p_2)

3/7* m_(p_2) = -1

▼

Solve for m_(p_2)

MultEqn

LHS * 7=RHS* 7

3m_(p_2) = -7

DivEqn

.LHS /3.=.RHS /3.

m_(p_2)=-7/3

MoveNegNumToFrac

Put minus sign in front of fraction

m_(p_2)=-7/3

The slope of the perpendicular bisector of DF is - 73. y=-7/3x+b Once again, to complete its equation, we can substitute the coordinates of its midpoint H( -11/2, 1-7/2).

y=-7/3x+b

SubstituteII

x= -11/2, y= -7/2

-7/2=-7/3( -11/2)+b

▼

Solve for b

MultNegNegOnePar

- a(- b)=a* b

-7/2=77/6+b

SubEqn

LHS-77/6=RHS-77/6

-7/2-77/6=b

ExpandFrac

a/b=a * 3/b * 3

-21/6-77/6=b

SubFrac

Subtract fractions

-98/6=b

ReduceFrac

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

-49/3=b

RearrangeEqn

Rearrange equation

b=-49/3

Therefore, the equation of the perpendicular bisector of DF is y=- 37x- 493.

Finding the Circumcenter

We want to find the intersection of the two perpendicular bisectors that we found. To do it, we need to solve the system of their equations. y=x y=-7/3x-49/3 We can solve it by substituting y=x in the second equation.

y=x & (I) y=-7/3x-49/3 & (II)

Substitute

(II): y= x

y=x x=-7/3x-49/3

▼

(II): Solve for x

AddEqn

(II): LHS+7/3x=RHS+7/3x

y=x x+7/3x=-49/3

OneToFrac

(II): Rewrite 1 as 3/3

y=x 3/3x+7/3x=-49/3

AddFrac

(II): Add fractions

y=x 10/3x=-49/3

MultEqn

(II): LHS * 3=RHS* 3

y=x 10x=-49

DivEqn

(II): .LHS /10.=.RHS /10.

y=x x=-49/10

Substitute

(I): x= -49/10

y= - 4910 x=- 4910

Therefore, the coordinates of the circumcenter are (- 4910,- 4910).