Question

A equação da reta que passa pelos pontos (-1,-2) e (5,2)

Resposta com calculo por favor !

Answer 1

Boa tarde Ligia!

Solução!

Vamos determinar o coeficiente angular da reta com essa formula.



$A(-1,-2)\\\\\ B(5,2)\\\\\\\ m= \dfrac{yB-yA}{xB-xA}\\\\\\\ m= \dfrac{2-(-2)}{5-(-1)}\\\\\\\ m= \dfrac{2+2}{5+1}\\\\\\\ m=\dfrac{4}{6}\\\\\\\ \boxed{m= \frac{2}{3} }$

Equação da reta!


$A(-1,-2)\\\\\ m= \dfrac{2}{3} \\\\\\ y-yA=m(x-xA)\\\\\\ y+2= \dfrac{2}{3} (x+1)\\\\\\ y+2= \dfrac{2x}{3}+ \dfrac{2}{3}\\\\\\ 3y+6=2x+2\\\\\\ 3y=2x+2-6\\\\\\ 3y=2x-4\\\\\ \boxed{y= \dfrac{2x-4}{3}}$

Boa tarde!
Bons estudos!

Answer 2

Oi Ligia

f(x) = ax + b

f(-1) = -a + b = -2
f(5) = 5a + b = 2

5a - (-a) = 2 - (-2)

5a + a = 2 + 2
6a = 4
a = 4/6 = 2/3

-2/3 + b = -6/3
b = -6/3 + 2/3 =-4/3

f(x) = (2x - 4)/3

equação reduzida
y = (2x - 4)/3

equação geral
2x - 3y - 4 = 0 

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