Lengths of medians from coordinates

There is a triangle ABC: A [-6.6; 1.2], B [3.4; -5.6], C [2.8; 4.2]. Calculate the lengths of its medians.

Final Answer:

t₁ = 9.8843
t₂ = 9.8478
t₃ = 7.7666

A = (- 6.6, 1.2) = (- 6.6, 1.2) B = (3.4, - 5.6) = (3.4, - 5.6) C = (2.8, 4.2) = (2.8, 4.2) T_{x} = \frac{A _{x} + B _{x} + C _{x}}{3} = \frac{( - 6 . 6 ) + 3 . 4 + 2 . 8}{3} = - \frac{2}{1 5} ≐ - 0.1333 T_{y} = \frac{A _{y} + B _{y} + C _{y}}{3} = \frac{1 . 2 + ( - 5 . 6 ) + 4 . 2}{3} = - \frac{1}{1 5} ≐ - 0.0667 \frac{A T}{T X} = 2 : 1 t_{1} = ∣ A T ∣ + ∣ T X ∣ A T = (T_{x} - A_{x})^{2} + (T_{y} - A_{y})^{2} = ((- 0.1333) - (- 6.6))^{2} + ((- 0.0667) - 1.2)^{2} ≐ 6.5896 t_{1} = \frac{3}{2} \cdot A T = \frac{3}{2} \cdot 6.5896 = 9.8843

B T = (T_{x} - B_{x})^{2} + (T_{y} - B_{y})^{2} = ((- 0.1333) - 3.4)^{2} + ((- 0.0667) - (- 5.6))^{2} ≐ 6.5652 t_{2} = \frac{3}{2} \cdot B T = \frac{3}{2} \cdot 6.5652 = 9.8478

C T = (T_{x} - C_{x})^{2} + (T_{y} - C_{y})^{2} = ((- 0.1333) - 2.8)^{2} + ((- 0.0667) - 4.2)^{2} ≐ 5.1777 t_{3} = \frac{3}{2} \cdot C T = \frac{3}{2} \cdot 5.1777 = 7.7666

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