Question

qual a medida da mediana AM cujos vértices é A(-1,2) B(-2,0) C(-1,-3) ?

Answer 1

O ponto $M\left(x_{_{M}},\,y_{_{M}} \right )$ é o ponto médio do segmento $\overline{BC}$. Encontrando as coordenadas do ponto $M$:

$x_{_{M}}=\dfrac{x_{_{B}}+x_{_{C}}}{2}\\ \\ x_{_{M}}=\dfrac{-2-1}{2}\\ \\ x_{_{M}}=-\dfrac{3}{2}\\ \\ \\ y_{_{M}}=\dfrac{y_{_{B}}+y_{_{C}}}{2}\\ \\ y_{_{M}}=\dfrac{0-3}{2}\\ \\ y_{_{M}}=-\dfrac{3}{2}$


O ponto médio do segmento $\overline{BC}$ é $M\left(-\dfrac{3}{2},\,-\dfrac{3}{2} \right )$.


A medida da mediana $AM$ é a distância entre os pontos $A$ e $M$:

$d_{AM}=\sqrt{\left(x_{_{M}}-x_{_{A}} \right )^{2}+\left(y_{_{M}}-y_{_{A}} \right )^{2}}\\ \\ d_{AM}=\sqrt{\left(-\dfrac{3}{2}-\left(-1 \right ) \right )^{2}+\left(-\dfrac{3}{2}-2 \right )^{2}}\\ \\ d_{AM}=\sqrt{\left(-\dfrac{3}{2}+1 \right )^{2}+\left(-\dfrac{3}{2}-2 \right )^{2}}\\ \\ d_{AM}=\sqrt{\left(\dfrac{-3+2}{2} \right )^{2}+\left(\dfrac{-3-4}{2} \right )^{2}}\\ \\ d_{AM}=\sqrt{\left(\dfrac{-1}{2} \right )^{2}+\left(\dfrac{-7}{2} \right )^{2}}\\ \\ d_{AM}=\sqrt{\dfrac{1}{4}+\dfrac{49}{4}}\\ \\ d_{AM}=\sqrt{\dfrac{50}{4}}\\ \\ d_{AM}=\dfrac{\sqrt{50}}{\sqrt{4}}\\ \\ d_{AM}=\dfrac{\sqrt{5^{2}\cdot 2}}{2}\\ \\ \boxed{\begin{array}{c} d_{AM}=\dfrac{5\sqrt{2}}{2}\text{ u.c.} \end{array}}$