Question

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

Answer 1

$Sejam\\\\\bullet\ M_1,\ o\ ponto\ m\'edio\ dos\ pontos\ A\ e\ B;e\\\bullet\ M_2,\ o\ ponto\ m\'edio\ dos\ pontos\ A\ e\ C$

$Dados\ dois\ pontos\ quaisquer\ A(x_1,y_1)\ e\ B(x_2,y_2)\\\\Ent\~ao$

$O\ ponto\ m\'edio\ entre\ desses\ dois\ pontos\ \'e\ dado\ pela\ f\'ormula$

$\boxed{xM= \dfrac{x_1+x_2}{2}\ e\ yM= \dfrac{y_1+y_2}{2}}}$

$\bullet\ Calculando\ M_1\\\\xM_1= \dfrac{2+(-3)}{2}\\\\\\xM_1= -\dfrac{1}{2}$

$yM_1= \dfrac{5+2}{2}\\\\\\yM_1= \dfrac{7}{2}$

$\bullet\ Calculando\ M_2\\\\xM_2= \dfrac{2+(-1)}{2}\\\\\\xM_2= \dfrac{1}{2}$

$yM_2= \dfrac{5+(-4)}{2}\\\\\\yM_2= \dfrac{1}{2}$

$Equa\c{c}\~ao\ da\ reta\ \'e\ dada\ pela\ f\'ormula\\\\y-y_O=m\cdot(x-x_O),\ onde\ m\ \'e\ o\ coeficiente\ angular\ da\ reta\\\\\\m= \dfrac{y_2-y_1}{x_2-x_1}$

$M_1\bigg(-\dfrac{1}{2},\dfrac{7}{2}\bigg) \ e\ M2\bigg( \dfrac{1}{2},\dfrac{1}{2} \bigg)\\\\\\\bullet\ Coeficiente\ angular$

$m = \dfrac{ \dfrac{1}{2}- \dfrac{7}{2}}{ \dfrac{1}{2}-\bigg( -\dfrac{1}{2}\bigg)}\\\\\\m= \dfrac{-\dfrac{6}{2}}{ \dfrac{2}{2} }\\\\\\m= -3\\\\\\y- \dfrac{1}{2}=-3\cdot\bigg(x- \dfrac{1}{2}\bigg)\\\\\\y- \dfrac{1}{2}=-3x+ \dfrac{3}{2}\\\\\\2y-1=-6x+3\\\\6x+2y-4=0\\\\Simplificando\ a\ equa\c{c}\~ao\ encontrada\\\\\\temos\\\\\\\boxed{3x+y-2=0}$

Para esboçar o gráfico, temos

$3x+y-2=0\\\\Logo, \\\\y = -3x+2$

Atribuindo valores para x, encontramos valores para y, veja o anexo I.

Marcando os pares ordenados no Sistema de Coordenadas Cartesiano, temos o gráfico da função encontrada, veja o anexo II.