Question

Dados os pontos A ( 2,5) , B (-3,2),C ( -1,-4), ache a equação geral da reta que passa pelos pontos médios de AB e AC.

Answer 1

Ponto médio de AB:

$P_{1}\left(\dfrac{x_{A}+x_{B}}{2},\dfrac{y_{A}+y_{B}}{2}\right)=\left(\dfrac{2-3}{2},\dfrac{5+2}{2}\right)=\left(-\dfrac{1}{2},\dfrac{7}{2}\right)$

Ponto médio de AC:

$P_{2}\left(\dfrac{x_{A}+x_{C}}{2},\dfrac{y_{A}+y_{C}}{2}\right)=\left(\dfrac{2-1}{2},\dfrac{5-4}{2}\right)=\left(\dfrac{1}{2},\dfrac{1}{2}\right)$
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Precisamos achar a equação da reta que passa por P₁ e P₂

Achando o coeficiente angular da reta:

$m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}=\dfrac{(\frac{1}{2})-(\frac{7}{2})}{(\frac{1}{2})-(-\frac{1}{2})}=\dfrac{(-\frac{6}{2})}{1}=-\dfrac{6}{2}=-3$

Achando a equação da reta:

$y=y_{1}+m\cdot(x-x_{1})~~~~ou~~~~y=y_{2}+m\cdot(x-x_{2})\\\\\\y=\dfrac{1}{2}-3\cdot\left(x-\dfrac{1}{2}\right)\\\\\\y=\dfrac{1}{2}-3x+\dfrac{3}{2}\\\\\\y=-3x+\dfrac{1+3}{2}\\\\\\\boxed{\boxed{y=-3x+2}}$