Explanation Let the triangle P N Q with vertices P - 2 a , 0 , N 2 a , 0 , and Q 0 , 2 b .is shown in the above figure. Let A J - , B K - , C H - be the perpendicular bisectors of the triangle P N Q . 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 l 1 ⊥ l 2 , then m 1 . m 2 = - 1 or m 2 = - 1 m 1 Since P N - is the horizontal line, slope is zero. (i.e.) m P N - = 0 Obviously the altitude C H - is vertical and it has the equation, x = 0 The slope of the line N Q - with vertices N 2 a , 0 , and Q 0 , 2 b is as follows. Using the slope formula and choosing x 1 = 2 a , x 2 = 0 , y 1 = 0 , and y 2 = 2 b . m N Q - = 2 b - 0 0 - 2 a m N Q - = 2 b - 2 a m N Q - = b - a Thus the slope of the altitude A J - which is perpendicular to the line N Q - is m A J - = a b To find any point on the line A J - , calculate the midpoint since the line is a perpendicular bisector of line N Q - . The midpoint A of the line N Q - that contains N 2 a , 0 , and Q 0 , 2 b is x , y = x 1 + x 2 2 , y 1 + y 2 2 x , y = 2 a + 0 2 , 0 + 2 b 2 x , y = a , b The equation of the line A J - that contains the point a , b with slope a b is as follows: y - b = a b x - a y - b = a x b - a 2 b y = a x b - a 2 b + b y = a b x + b 2 - a 2 b The intersection of these perpendicular bisector is found by solving the two perpendicular line equation. x = 0 and y = a b x + b 2 - a 2 b On solving, y = a b 0 + b 2 - a 2 b y = b 2 - a 2 b The coordinates of circumcenter x , y is 0 , b 2 - a 2 b

Elementary Geometry For College Students, 7e

7th Edition

ISBN: 9781337614085

Author: Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher: Cengage,

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Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

Chapter 10.5, Problem 46E

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

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Chapter 10.5, Problem 45E

Chapter 10.5, Problem 47E

Chapter 10 Solutions

Elementary Geometry For College Students, 7e

Ch. 10.1 - Plot and then label the points A0,-3, B3,-4, C5,...Ch. 10.1 - Give the coordinates of each point A, B, C, D, and...Ch. 10.1 - Find the distance between each pair of points: a...Ch. 10.1 - If the distance between -2, 3 and -2,a is 5 units,...Ch. 10.1 - If the distance between b,3 and 7, 3 is 3.5 units,...Ch. 10.1 - Prob. 6E Ch. 10.1 - Find the distance between each pair of points: a...Ch. 10.1 - Find the distance between each pair of points: a...Ch. 10.1 - Find the midpoint of the line segment that joins...Ch. 10.1 - Prob. 10E

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To find: The perpendicular bisectors of the sides of a triangle is concurrent.

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Chapter 10 Solutions

To find:

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