show that (2,1) is the centre of the circumcircle of the triangle whose vertices are (-3,-9),(13,-1) and (-9,3).

Question

show that (2,1) is the centre of the circumcircle of the triangle whose vertices are (-3,-9),(13,-1) and (-9,3).

in progress 0
Reagan 18 hours 2021-10-13T09:48:55+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-13T09:50:10+00:00

    Given : (2,1) is the centre of the circumcircle of the triangle whose vertices are (-3,-9),(13,-1) and (-9,3).

    To find : verify

    Solution :

    centre of the circumcircle of the triangle is at equal distance from all the vertex of triangle

    centre of the circumcircle ( 2 ,1 )

    Vertex   are (-3,-9),(13,-1) and (-9,3).

    Distance of  ( 2 ,1 )  from  (-3,-9)

    = √(2 – (-3))² + (1 – (-9))²

    = √5² + 10²

    = √25 + 100

    = √125

    = 5√5

    Distance of  ( 2 ,1 )  from  (13,-1)

    = √(2 – (13))² + (1 – (-1))²

    = √(-11)² + (2)²

    = √121 + 4

    = √125

    = 5√5

    Distance of  ( 2 ,1 )  from  (-9,3)

    = √(2 – (-9))² + (1 – (3))²

    = √11² + (-2)²

    = √121 + 4

    = √125

    = 5√5

    (2,1) is at equal distance 5√5 from all the vertex hence

    (2,1) is the centre of the circumcircle of the triangle whose vertices are (-3,-9),(13,-1) and (-9,3).

    Learn more:

    Which of the following is a characteristic of the circumcenter of a …

    https://brainly.in/question/15038473

    If (cosalpha,sinalpha,0),(cosbeta,sinbeta,0),cosgama,singama,0 …

    https://brainly.in/question/5972494

    0
    2021-10-13T09:50:11+00:00

    Here is the explanation

    Hope you liked it

Leave an answer

Browse

14:4+1-6*5-7*14:3+5 = ? ( )