   Chapter 10.5, Problem 46E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

In Exercises 45 and 46, complete an analytic proof for each theorem.The perpendicular bisectors of the sides of a triangle are concurrent.

To determine

To find:

The perpendicular bisectors of the sides of a triangle is concurrent.

Explanation

Let the triangle PNQ with vertices P-2a, 0, N2a, 0, and Q0, 2b.is shown in the above figure.

Let AJ-, BK-, CH- be the perpendicular bisectors of the triangle PNQ.

The point of intersection of the perpendicular bisectors is the circumcentre.

By theorem,

If two lines are perpendicular, then the product of their slopes is -1 or one slope is negative reciprocal of the other slope.

(i.e) If l1l2, then m1.m2=-1 or m2=-1m1

Since PN- is the horizontal line, slope is zero.

(i.e.) mPN-=0

Obviously the altitude CH- is vertical and it has the equation,

x=0

The slope of the line NQ- with vertices N2a, 0, and Q0, 2b is as follows.

Using the slope formula and choosing x1=2a, x2=0, y1=0, and y2=2b.

mNQ-=2b-00-2a

mNQ-=2b-2a

mNQ-=b-a

Thus the slope of the altitude AJ- which is perpendicular to the line NQ- is

mAJ-=ab

To find any point on the line AJ-, calculate the midpoint since the line is a perpendicular bisector of line NQ-.

The midpoint A of the line NQ- that contains N2a, 0, and Q0, 2b is

x, y=x1+x22, y1+y22

x, y=2a+02, 0+2b2

x, y=a, b

The equation of the line AJ- that contains the point a,b with slope ab is as follows:

y-b=abx-a

y-b=axb-a2b

y=axb-a2b+b

y=abx+b2-a2b

The intersection of these perpendicular bisector is found by solving the two perpendicular line equation.

x=0 and y=abx+b2-a2b

On solving,

y=ab0+b2-a2b

y=b2-a2b

The coordinates of circumcenter x, y is 0, b2-a2b

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