Question

12) Os pontos A = (0,0), B = (3,7) e C = (5, -1) são vértices de um triângulo. O comprimento da mediana AM vale:

Answer 1

Boa tarde!

Solução!

Primeiro vamos determinar a coordenada do ponto médio do segmento BC.

$m(BC)= \left (\dfrac{xB+Xc}{2}, \dfrac{yB+yC}{2} \right )\\\\\\\ m(BC)= \left (\dfrac{3+5}{2}, \dfrac{7-1}{2} \right )\\\\\\\ m(BC)= \left (\dfrac{8}{2}, \dfrac{6}{2} \right )\\\\\\\ \boxed{m(BC)= (4, 3 )}$

Agora vamos calcular a distância do vértice A ate o ponto médio.

$m(BC)= (4, 3 )\\\\\\ A(0,0)\\\\\\ d(A,m)= \sqrt{(xA-xm)^{2} +(yA-ym)^{2} }\\\\\\\\ d(A,m)= \sqrt{(0-4)^{2} +(0-3)^{2} }\\\\\\\\ d(A,m)= \sqrt{(-4)^{2} +(-3)^{2} }\\\\\\\\ d(A,m)= \sqrt{16 +9}\\\\\\\\ d(A,m)= \sqrt{25}\\\\\\\\ d(A,m)=5\\\\\\\\\\ Resposta:Comprimento~~da~~mediana~~igual~~5$

Boa tarde!
Bons estudos!