Exercise 39 Page 558

Since B' is a reflection of B, then c is a perpendicular bisector of BB'. Let's show the image, preimage, and line c in a diagram.

To find the equation of line c, we need to know its slope and y-intercept. We know that line c runs perpendicular to BB' so let's start by finding the slope of this segment.

m = y_2 - y_1/x_2 - x_1

SubstitutePoints

Substitute ( 3,2) & ( 1,4)

m = 2 - 4/3 - 1

▼

Simplify right-hand side

SubTerms

Subtract terms

m = - 2/2

CalcQuot

Calculate quotient

m = - 1

The slope of BB' is - 1. Because perpendicular lines have slopes that are opposite reciprocals, we know that the slope of line c is m=1. So far we have the equation y=1x+b To calculate the y-intercept, we need a point on the line. Let's use the fact that BB' is a perpendicular bisector. This means the intersection of line c and BB' is the midpoint of BB'. Thus, we can use the Midpoint Formula to find this point.

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

SubstitutePoints

Substitute ( 1,4) & ( 3,2)

M(1+ 3/2,4+ 2/2)

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Simplify

AddTerms

Add terms

M(4/2,6/2)

CalcQuot

Calculate quotient

M(2,3)

Now we have enough information to calculate the y-intercept.

y=x+b

SubstituteII

x= 2, y= 3

3= 2+b

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Solve for b

SubEqn

LHS-2=RHS-2

1=b

RearrangeEqn

Rearrange equation

b=1

Line c has the equation y=x+1. Let's show it as well.