Triangle midpoints

Determine coordinates of triangle ABC vertices if we know triangle sides midpoints SAB [0;3] SBC [1;6] SAC [4;5], its sides AB, BC, AC.

Final Answer:

Ax = 3
Ay = 2
Bx = -3
By = 4
Cx = 5
Cy = 8

Q = (0, 3) W = (1, 6) T = (4, 5) \frac{A _{x} + B _{x}}{2} = Q_{x} \frac{A _{y} + B _{y}}{2} = Q_{y} \frac{B _{x} + C _{x}}{2} = W_{x} \frac{B _{y} + C _{y}}{2} = W_{y} \frac{C _{x} + A _{x}}{2} = T_{x} \frac{C _{y} + A _{y}}{2} = T_{y} A_{x} + B_{x} = 2 \cdot Q_{x} A_{y} + B_{y} = 2 \cdot Q_{y} B_{x} + C_{x} = 2 \cdot W_{x} B_{y} + C_{y} = 2 \cdot W_{y} C_{x} + A_{x} = 2 \cdot T_{x} C_{y} + A_{y} = 2 \cdot T_{y} A_{x} + (2 \cdot W_{x} - C_{x}) = 2 \cdot Q_{x} A_{x} + (2 \cdot W_{x} - (2 \cdot T_{x} - A_{x})) = 2 \cdot Q_{x} A_{x} + 2 \cdot W_{x} - 2 \cdot T_{x} + A_{x} = 2 \cdot Q_{x} A x = A_{x} = \frac{2 \cdot Q _{x} - 2 \cdot W _{x} + 2 \cdot T _{x}}{2} = \frac{2 \cdot 0 - 2 \cdot 1 + 2 \cdot 4}{2} = 3

A y = A_{y} = \frac{2 \cdot Q _{y} - 2 \cdot W _{y} + 2 \cdot T _{y}}{2} = \frac{2 \cdot 3 - 2 \cdot 6 + 2 \cdot 5}{2} = 2

B x = B_{x} = \frac{2 \cdot W _{x} - 2 \cdot T _{x} + 2 \cdot Q _{x}}{2} = \frac{2 \cdot 1 - 2 \cdot 4 + 2 \cdot 0}{2} = - 3

B y = B_{y} = \frac{2 \cdot W _{y} - 2 \cdot T _{y} + 2 \cdot Q _{y}}{2} = \frac{2 \cdot 6 - 2 \cdot 5 + 2 \cdot 3}{2} = 4

C x = C_{x} = \frac{2 \cdot T _{x} - 2 \cdot Q _{x} + 2 \cdot W _{x}}{2} = \frac{2 \cdot 4 - 2 \cdot 0 + 2 \cdot 1}{2} = 5

C y = C_{y} = \frac{2 \cdot T _{y} - 2 \cdot Q _{y} + 2 \cdot W _{y}}{2} = \frac{2 \cdot 5 - 2 \cdot 3 + 2 \cdot 6}{2} = 8

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