In this page solution of missingvertex question7 we are going to see he detailed solution of the first question.

**Question 1 :**

**Find the ****missing vertex of triangle whose two vertices are (-3,-9) (-1,6) and its centroid is (-1/3 , 2 )**

**Question 2 :**

**Find the ****missing vertex of triangle whose two vertices are (-3,-9) **** (5,-8)** and its centroid is (5/3 , -8/3 )

**Question 3 :**

**Find the ****missing vertex of triangle whose two vertices are (4,5) **** (4,2)** and its centroid is (2 , 3 )

**Question 4 :**

**Find the ****missing vertex of triangle whose two vertices are ****(3,1) (2,2)**** and its centroid is (7/3 , 1 )**

**Question 5 :**

**Find the ****missing vertex of triangle whose two vertices are ****(3,1) (0,4)**** and its centroid is (0 , 2 )**

**Question 6 :**

**Find the ****missing vertex of triangle whose two vertices are ****(1,10) (-7,2)**** and its centroid is (-3 , 19/3 )**

**Question 7 :**

**Find the ****missing vertex of triangle whose two vertices are ****(-1,-3) (2,1)**** and its centroid is (1 , -2 )**

**Solution:**

**Centroid of a triangle = (x****₁+x****₂+x****₃)/3, (y****₁+y****₂+y****₃)/3**

let A,B and C be the vertices of triangle ABC. Here A (-1, -3), B (2,1) and let us take the third missing vertex be (a,b).

x₁ = -1 x₂ = 2 x₃ = a

y₁ = -3 y₂ = 1 y₃ = b

(1, -2) = [ (-1) + 2 + a ]/3 , [ (-3) + 1 + b ]/3

(1 , -2) = [-1 + 2 + a ]/3 , [ -3 + 1 + b ]/3

(1, -2) = [1 + a ]/3 , [-2 + b ]/3

Equating x and y coordinates

[1 + a ]/3 = 1 [ -2 + b ]/3 = -2

1 + a = 1 x 3 - 2 + b = -2 x 3

1 + a = 3 -2 + b = -6

a = 3 - 1 b = -6 + 2

a = 2 b = -4

Therefore the third vertex is (2,-4).

**Question 8 :**

**Find the ****missing vertex of triangle whose two vertices are ****(1,1) (2,3)**** and its centroid is (-1/3 , 2 )**