Question

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

Answer 1

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

`(x_n,y_m) = (((x_1 + y_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) (3, -2)`,

`M_(BC) (-3, 1) and M_(CA) (4, -3)`.

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

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

Equating the individual components we get,

`x_A + x_B = 6`

`y_A + y_B = -4`

Using the midpoint of another side we have,

`(-3,1) = (((x_B + x_C)/2)","((y_B + y_C)/2))`

Equating the individual components we get,

`x_B + x_C = -6`

`y_B + y_C = 2`

Using the midpoint of the last side we have,

`(4, -3) = (((x_A + x_C)/2)"," ((y_A + y_C)/2))`

Equating the individual components we get,

`x_A + x_C = 8`

`y_A + y_C = -6`

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 = 6 - 6 + 8`

`2(x_A + x_B + x_C) = 8`

`x_A + x_B + x_C = 4`

Substituting `x_B + x_C = -6`in the above equation we have

`x_A + x_B + x_C = 4`

`x_A - 6 = 4`

`x_A = 10`

Therefore,

`x_A + x_C = 8`

`x_C = 8 - 10`

`x_C = -2`

And

`x_A + x_B = 6`

`x_B = 6 - 10`

`x_B = -4`

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 = -4 + 2 - 6`

`2(y_A + y_B + y_C) = -8`

`y_A + y_B + y_C = -4`

Substituting `y_B + y_C = 2` in the above equation we have,

`y_A + y_B + y_C = 4`

`y_A + 2 = -4`

`y_A = -6`

Therefore,

`y_A + y_C = -6`

`y_C = -6 + 6`

`y_C  = 0`

And

`y_A + y_B = -4`

`y_B = -4 + 6`

`y_B = 2`

Therefore the co-ordinates of the three vertices of the triangle are A(10, 6), B(-4, 2), C(-2, 0)