Question

If (-3, 2), (1, -2) and (5, 6) are the midpoints of the sides of a triangle, find the coordinates of the vertices of the triangle. 

Answer 1

let A(x₁, y₁ ), B(x₂, y₂) and O(x₃ , y₃) be the coordinates of the vertices of Δ ABC. 

D is the midpoint of AB <

D(-3,2) = D `(("x"_1 + "x"_2)/2 , ("y"_1 + "y"_2)/2)`

`("x"_1 + "x"_2)/2 = -3 , ("y"_1 + "y"_2)/2`

X₁ + X₂ = -6 - - - (1)                  Y₁ + Y₂ = 4   .......(2)

Similarly 

X₂ + X₃ = 2 - - - (3)                  Y2 + Y3 = -4   ......(4)

X₁ + X₃ = 10 - - - ( 5)                Y₁ + Y₃ = 12   .......(6)

Adding (1), (3) and (5)  

2(x₁ + x₂ + x,) = 6 

x₁ + x₂ + x₃ = 3 

-6 + x₃ = 3

x₃ = 9

From (3)

x₂ + 9 = 2

x₂ = 9

From (3) 

X₂ + 9 = 2 

X₂ = -7 

From (5) 

x₁ +9=10 

x₁ = 1 

Adding (2), ( 4) and (6) 

2(y₁ + Y₂+y₃)= 12 

Y₁+ Y₂ + Y₃ = 6 

4 + y₃ = 6 

Y₃ = 2 

from(4) 

y₂ + 2 = -4 

Y₂ = -6 

from(6) 

Y₁+2= 12 

Y₁ = 10

The coordinates of the vertices of  Δ ABC are (9,2), (1, 10) and (-7,-6).