#### Question

If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, —3) and (3, 4), find the vertices of the triangle.

#### Solution

The coordinates of the midpoint `(x_m,y_m)` between two points `(x_1,y_1)` and `(x_2, y_2)` is given by,

`(x_m,y_m) = (((x_1 + x_2)/2)"," ((y_1 + y_2)/2))`

Let the three vertices of the triangle be `A(x_A,y_A),B(x_B, y_B)` and `C(x_C, y_C)`

The three midpoints are given. Let these points be `M_(AB) (1,1), M_(BC) (2, -3)`and `M_(CA) (3,4)`

Let us now equate these points using the earlier mentioned formula,

`(1,1) = (((x_A + x_B)/2)","((y_A + y_B)/2))`

Equating the individual components we get,

`x_A + x_B = 2`

`y_A + y_B = 2`

Using the midpoint of another side we have,

`(2,-3) = (((x_B + x_C)/2)","((y_B + y_C)/2))`

Equating the individual components we get,

`x_A + x_C= 6`

`y_A + y_C= 8`

Adding up all the three equations which have variable ‘*x*’ alone we have,

`x_A + x_B + x_B + x_C + x_A + x_C = 2 + 4 + 6`

`2(x_A + x_B + x_C) = 12`

`x_A + x_B +x_C = 6`

Substituting `x_B + x_C = 4` in the above equation we have,

`x_A + x_B + x_C = 6`

`x_A + 4 = 6`

`x_A = 2`

Therefore,

`x_A + x_C= 6`

`x_C = 6 - 2`

`x_C = 4`

And

`x_A + x_B 2`

x_b = 2 - 2

`x_B = 0`

Adding up all the three equations which have variable ‘*y*’ alone we have,

`y_A + y_B + y_B + y_C + y_A + y_C= 2 - 6 + 8`

`2(y_A + y_B + y_C) = 4`

`y_A + y_B + y_C = 2`

Substituting `y_B + y_C = -6` in the above equation we have,

`y_A + y_B + y_C = 2`

`y_A - 6 = 2`

`y_A = 8`

Therefore,

`y_A + y_C = 8`

`y_C = 8 - 8`

`y_C = 0`

And

`y_A + y_B = 2`

`y_B = 2 - 8`

`y_B = -6`

Therefore the co-ordinates of the three vertices of the triangle are A(2,8), B(0, -6), C(4, 0)