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 (x₁, y₁) and (x₂, y₂) 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, Co-ordinates of the three vertices of the triangle are A (4, 5), B (2, 3), C (6, 9).