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.


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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

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