   Chapter 10.5, Problem 44E 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

For isosceles ∆ P N Q , the vertices are P - 2 a ,   0 , N 2 a ,   0 , and Q 0 ,   2 b . In terms of a , and b , find the coordinates of the circumcenter of ∆ P N Q . (The circumcenter is the point of concurrence for the perpendicular bisectors of the sides of a triangle.)

To determine

To find:

The coordinates of the circumcenter of an isosceles PNQ in terms of a, and b with the vertices P-2a, 0, N2a, 0, and Q0, 2b.

Explanation

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-

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