The orthocenter and circumcentre of triangle ABC are (1 2) and (2 4) respectively.If the base BC has the equation 2x-y-3=0 .The radius of the circumcircle of triangle ABC is 1.sqrt((57)/(5)) 2.sqrt((59)/(5)) 3.sqrt((61)/(5)) 4.sqrt((63)/(5))..
Answers
Given : The orthocenter and circumcentre of triangle ABC are (1 2) and (2 4) respectively. Base BC has the equation 2x-y-3=0
To find : The radius of the circumcircle of triangle ABC
Solution:
The centroid divides the distance from the orthocentre to the circumcentre in the ratio 2:1.
orthocentre = ( 1 . 2)
circumcentre = ( 2, 4)
Centroid = (2 * 2 + 1 *1)/(2 + 1) . ( 2 * 4 + 1 * 2)/(2 + 1)
= 5/3 , 10/3
circumcentre is on perpendicular bisector
perpendicular to 2x - y - 3 = 0
hence slope = -1/2 and passes through ( 2 , 4)
=> y - 4 = (-1/2)(x - 2)
=> 2y - 8 = -x + 2
=> x + 2y - 10 = 0
on Solving :
mid point of BC = x = 16/5 , y = 17/5 let say D = ( 16/5 , 17/5)
Centroid Divided median in 2 : 1 ratio
let say A = ( h , k)
=> 5/3 = ( 2 * (16/5) + 1*h)/(2 + 1) & 10/3 = (2*(17/5) + 1.k)/(2 + 1)
=> h =5 - 32/5 = -7/5 & k = 10 - 34/5 = 16/5
A = ( - 7/5 , 16/5)
Circum center = Distance between A = ( - 7/5 , 16/5) & circumcentre ( 2 , 4)
AC = √ (2 + 7/5)² + (4 - 16/5)² = √(17² + 4²) / 5 = √(289 + 16) /5
= √305 / 5
= √(61/5)
radius of the circumcircle of triangle ABC is √(61/5)
option 3 is correct
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