If-the-points-3-5-4-2-and-6-2-are-the-vertices-of-a-triangle-i-Find-the-equation-of-the-perpendicular-bisector-of-the-sides-ii-Find-the-coordinate-of-the-circumcenter-The-circumcen

Question Number 43944 by Tawa1 last updated on 18/Sep/18

If the points (- 3, 5), (4, - 2) and (6, 2) are the vertices of a triangle .

(i) Find the equation of the perpendicular bisector of the sides

(ii) Find the coordinate of the circumcenter . (The circumcenter of a

triangle is the point of intersection of the perpendicular bisector of

the side

(iii) Find the radius of the circumcircle .

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Sep/18

A (- 3, 5) B (4, - 2) C (6, 2)

A B = (y - 5) = \frac{- 2 - 5}{4 + 3} (x + 3)

7 y - 35 = - 7 x - 21

7 y + 7 x = 14

m i d o f A B \frac{- 3 + 4}{2}, \frac{5 - 2}{2} (\frac{1}{2}, \frac{3}{2})

p e r p e n d i c u l a r t o A B a n d p a s s i n g t h r o u g h (\frac{1}{2}, \frac{3}{2}) i s

p e r p d n d i c u l R b i s e c t o r o f A B

n o w s l o p e o f A B c a l c u l a t i o n

7 y + 7 x = 14

y = - x + 2 m = - 1

m_{1} \times m_{2} = - 1

(- 1) \times m_{2} = - 1 m_{2} = 1

s o ⊥ b i s e c t o r o f A B i s (y - \frac{3}{2}) = 1 (x - \frac{1}{2})

2 y - 3 = 2 x - 1

2 y - 2 x = 2

y - x = 1

s i m i l a r w a y c a l c u l a t e o t h e r ⊥ b i s e c t o r

o f B C B (4, - 2) C (6, 2)

B C e q n i s y + 2 = \frac{2 + 2}{6 - 4} (x - 4)

2 y + 4 = 4 x - 16

2 y - 4 x + 20 = 0 y - 2 x + 10 = 0

s l o p e = 2

m_{1} \times m_{2} = - 1 2 \times m_{2} = - 1 m_{2} = \frac{- 1}{2}

m i d p o i n t o f B C {\frac{4 + 6}{2}}^{'} \frac{- 2 + 2}{2} = (5, 0)

e q n o f ⊥ b i s e c t o r o f B C i s

y - 0 = \frac{- 1}{2} (x - 5)

2 y + x - 5 = 0

i n t e r s e c t i n g p o i n t b e t w e e n y - x = 1

a n d 2 y + x - 5 = 0

i s t h e c e n t r e o f c i r c l e

y - x = 1

2 y + x = 5

3 y = 6 y = 2 x = 1

s o c e n t r e i f c i r c l e i s (1, 2)

e q n

{(x - 1)}^{2} + {(y - 2)}^{2} = r^{2}

r = d i s t a n c e b e t w e e n (1, 2) a n d (- 3, 5) \leftarrow A p o i n t

r = \sqrt{{(- 3 - 1)}^{2} + {(5 - 2)}^{2}} = \sqrt{16 + 9} = 5

s o e q n c i r c l e {(x - 1)}^{2} + {(y - 2)}^{2} = 5^{2}

a n o t h e r w a y

i i) e q n o f c i r c l e i s x^{2} + y^{2} + 2 g x + 2 f y + c = 0

p a s s e s t h r o u g h (- 3, 5) (4, - 2) a n d (6, 2)

9 + 25 - 6 g + 10 f + c = 0 \leftarrow e q n 1

16 + 4 + 8 g - 4 f + c = 0 \leftarrow e q n 2

36 + 4 + 12 g + 4 f + c = 0 \leftarrow e q n 3 f c i r c l e

34 - 6 g + 10 f + c = 0

20 + 8 g - 4 f + c = 0

40 + 12 g + 4 f + c = 0

e q n 1 - e q n 2

14 - 14 g + 14 f = 0 1 - g + f = 0

e q n 2 - e q n 3

- 20 - 4 g - 8 f = 0

5 + g + 2 f = 0

s o l v e

5 + g + 2 f = 0

1 - g + f = 0

6 + 3 f = 0 f = - 2 g = - 1

c e n t r e i s (- g, - f) = (1, 2)

p u t g = - 1 a n f f = - 2 i n e q n 1 t o [f i n d c

34 - 6 g + 10 f + c = 0

34 - 6 (- 1) + 10 (- 2) + c = 0

34 + 6 - 20 + c = 0

c = - 20

r a d i u s r = \sqrt{g^{2} + f^{2} - c}

r = \sqrt{1 + 4 + 20} = 5

e q n c i r c l e

{(x - 1)}^{2} + {(y - 2)}^{2} = 5^{2}

Commented by Tawa1 last updated on 18/Sep/18

God bless you sir . I really appreciate

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