Exercise 21 Page 307

We have to find the equation of the perpendicular bisector of the segment whose endpoints are U(-3,4) and V(9,8). We will do this in three steps.

1. Find the midpoint of the segment.
2. Find the slope of the perpendicular bisector.
3. Use the point-slope form to write the equation of the line.

Let's go for it!

Midpoint of the Segment

A segment bisector contains the midpoint of the segment. We can find its coordinates using the Midpoint Formula. M( x_1+x_2/2,y_1+y_2/2 )Let's substitute the given endpoints U(-3,4) and V(9,8) into the above formula.

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

SubstitutePoints

Substitute ( -3,4) & ( 9,8)

M( -3+ 9/2,4+ 8/2 )

AddTerms

Add terms

M( 6/2,12/2 )

CalcQuot

Calculate quotient

M(3,6)

Therefore, the coordinates of the midpoint are (3,6).

Slope of the Perpendicular Bisector

A perpendicular bisector is perpendicular to the segment through the midpoint. In order to find the slope of the bisector, we will first find the slope of the segment. To do so, we can use the Slope Formula. m = y_2-y_1/x_2-x_1 We can choose any two points on the segment. Let's use the endpoints U(-3,4) and V(9,8).

m = y_2-y_1/x_2-x_1

SubstitutePoints

Substitute ( -3,4) & ( 9,8)

m=8- 4/9-( -3)

SubNeg

a-(- b)=a+b

m=8-4/9+3

AddSubTerms

Add and subtract terms

m=4/12

ReduceFrac

a/b=.a /4./.b /4.

m=1/3

The slope of the segment is 13. Let m_p be the slope of the perpendicular bisector. The product of the slopes of two perpendicular lines is - 1. 1/3 * m_p = - 1 ⇔ m_p = - 3 The slope of the perpendicular bisector is -3.

Equation of the Perpendicular Bisector

Since we know a point and the slope of the bisector, we can use the point-slope form of a line to write its equation. y-y_1=m(x-x_1) Let's substitute the slope -3 and the midpoint ( 3, 6) as a point into the formula.

y-y_1=m(x-x_1)

SubstituteValues

Substitute values

y- 6= -3(x- 3)

Distr

Distribute -3

y-6=-3 x-3(-3)

MultNegNegOnePar

- a(- b)=a* b

y-6=-3 x+9

AddEqn

LHS+6=RHS+6

y=-3 x+15

The equation of the perpendicular bisector of the segment whose endpoints are U(-3,4) and V(9,8) is y=-3 x+15.