• Matéria: Matemática
  • Autor: jaquelinesouza15
  • Perguntado 9 anos atrás

me ajudem por favor nesse exercicio :Escreva a equacao geral da reta que passa pelos pontos indicados. Determine um ponto C que esteja na reta:B)A(5,2) e D(-1-6)

Respostas

respondido por: Anônimo
2
Boa noite! 

Solução!

A(5,2)\\\\
D(-1,-6)\\\\\\
1\º)~~Determinar~~o~~coeficiente~~angular~~que~~sera~~indicado\\\\por~~m\\\\\\
m= \dfrac{yD-yA}{xD-xA}\\\\\\\
m= \dfrac{-6-2}{-1-5}\\\\\\\ 
m= \dfrac{-8}{-6}\\\\\\\

\boxed{m= \dfrac{8}{6}}

Equação da reta!

A(5,2)\\\\\\
m= \dfrac{8}{6}\\\\\\\\
y-yA=m(x-xA)\\\\\
y-2= \dfrac{8}{6}(x-5)\\\\\\\ 
y-2= \dfrac{8x-40}{6}\\\\\\\
6y-12=8x-40\\\\\
-8x+6y-12+40=0.(-1)\\\\\\
\boxed{8x-6y-28=0}~~Eq~~da~~reta~~na~~forma~~geral.

Ponto C pertencente a reta!

C(x,y)\\\\\\

8x-6y-28=0\\\\\
8(2)-6(-2)-28=0\\\\\
16+12-28=0\\\\\
28-28=0\\\\\
0=0\\\\\
C(2,-2)

Boa noite!
Bons estudos!

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