Question 3 :
Find the missing vertex of triangle whose two vertices are (4,5) (4,2) and its centroid is (2 , 3 )
Solution:
Centroid of a triangle = (x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3
let A,B and C be the vertices of triangle ABC. Here A (4, 5), B (4,2) and let us take the third missing vertex be (a,b).
x₁ = 4 x₂ = 4 x₃ = a
y₁ = 5 y₂ = 2 y₃ = b
(2, 3) = [ 4 + 4 + a ]/3 , [ 5 + 2 + b ]/3
(2 , 3) = [8 + a ]/3 , [ 7 + b ]/3
(2 , 3) = [8 + a ]/3 , [7 + b ]/3
Equating x and y coordinates
[8 + a ]/3 = 2 [ 7 + b ]/3 = 3
8 + a = 2 x 3 7 + b = 3 x 3
8 + a = 6 7 + b = 9
a = 6 - 8 b = 9 - 7
a = -2 b = 2
Therefore the third vertex is (-2,2).
Question 4 :
Find the missing vertex of triangle whose two vertices are (3,1) (2,2) and its centroid is (7/3 , 1 )