Question

If the coordinates of the mid-points of the sides of a triangle are (3, 4) (4, 6) and (5, 7), find its vertices.

Answer 1

The co-ordinates of the midpoint `(x_m,y_m)` between two points `(x_1,y_1)` and (x_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)` (3,4), `M_(BC)` (4, 6)`and `M_(CA) (5, 7)`.

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

`(3,4) = (((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 = 8`

Using the midpoint of another side we have,

`(4,6) = (((x_B + x_C)/2)","((y_B + y_C)/2))`

Equating the individual components we get,

`x_B + x_C = 8`

`y_B + y_C = 12`

Using the midpoint of the last side we have,

`(5,7) = (((x_A + x_C)/2)","((y_A + y_C)/2))`

Equating the individual components we get,

`x_A + x_C = 10`

`y_A + y_C = 14`

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 + 8 + 10`

`2(x_A + x_B + x_C) = 24`

`x_A + x_B + x_C = 12`

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

`x_A + x_B + x_C = 12`

`x_A + 8 = 12`

`x_A = 4`

Therefore,

`x_A + x_C = 10`

`x_C = 10 - 4`

`x_C = 6`

And

`x_A + x_B = 6`

`x_B =   6 - 4`

`x_B = 2`

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 = 8 + 12 + 14`

`2(y_A + y_B + y_C) = 34`

`y_A + y_B + y_C = 17`

Substituting `y_B + y_C = 12` in the above equation we have

`y_A + y_B + y_C = 17`

`y_A + 12 = 17`

`y_A = 5`

Therefore

`y_A + y_C = 14`

`y_C = 14 - 5`

`y_C = 9

And

`y_A + y_B = 8`

`y_C = 14 - 5`

`y_C = 9`

And

`y_A + y_B = 8`

`y_B = 8 - 5`

`y_B = 3`

Therefore the co-ordinates of the three vertices of the triangle are A(4,5), B(2,3), C(6,9)