Question

como determinar a mediana de um segmento Am de um triângulo cujas vértices são A(0,-3) B(-7,-1) e C (2,-2)?

Answer 1

A mediana $\text{AM}$ tem extremidades nos pontos $\text{A}(0,-3)$ e $\text{M}(x_\text{M},y_\text{M})$, sendo $\text{M}$ o ponto médio de $\text{BC}$

Como $\text{B}(-7,-1)$ e $\text{C}(2,-2)$ as coordenadas de $\text{M}$ são:

$x_\text{M}=\dfrac{x_\text{B}+x_\text{C}}{2}=\dfrac{-7+2}{2}=-\dfrac{5}{2}$

$y_\text{M}=\dfrac{y_\text{B}+y_\text{C}}{2}=\dfrac{-1-2}{2}=-\dfrac{3}{2}$

Desse modo, $\text{M}(-\frac{5}{2},-\frac{3}{2})$ e a mediana $\text{AM}$ mede:

$\text{AM}=\sqrt{\left(0+\dfrac{5}{2})\right)^2+\left(-3+\dfrac{3}{2}\right)^2}$

$\text{AM}=\sqrt{\left(\dfrac{5}{2}\right)^2+\left(-\dfrac{3}{2}\right)^2}$

$\text{AM}=\sqrt{\dfrac{25}{4}+\dfrac{9}{4}}$

$\text{AM}=\sqrt{\dfrac{34}{4}$

$\text{AM}=\dfrac{\sqrt{34}}{2}$