Question

ache a equação geral da reta:
a)(3,4) e (6,5)
b)(-5,6) e (0,-2)
c)(2,3) e (4,8)

Answer 1

Explicação passo-a-passo:

a) $\sf A(3,4),~B(6,5)$

• Coeficiente angular

$\sf m=\dfrac{y_B-y_A}{x_B-x_A}$

$\sf m=\dfrac{5-4}{6-3}$

$\sf m=\dfrac{1}{3}$

• Equação da reta

$\sf y-y_0=m\cdot(x-x_0)$

$\sf y-4=\dfrac{1}{3}\cdot(x-3)$

$\sf 3y-12=x-3$

$\sf x-3y-3+12=0$

$\sf x-3y+9=0$

b) $\sf A(-5,6),~B(0,-2)$

• Coeficiente angular

$\sf m=\dfrac{y_B-y_A}{x_B-x_A}$

$\sf m=\dfrac{-2-6}{0-(-5)}$

$\sf m=\dfrac{-2-6}{0+5}$

$\sf m=\dfrac{-8}{5}$

• Equação da reta

$\sf y-y_0=m\cdot(x-x_0)$

$\sf y-(-2)=\dfrac{-8}{5}\cdot(x-0)$

$\sf y+2=\dfrac{-8}{5}\cdot(x-0)$

$\sf 5y+10=-8x$

$\sf 8x+5y+10=0$

c) $\sf A(2,3),~B(4,8)$

• Coeficiente angular

$\sf m=\dfrac{y_B-y_A}{x_B-x_A}$

$\sf m=\dfrac{8-3}{4-2}$

$\sf m=\dfrac{5}{2}$

• Equação da reta

$\sf y-y_0=m\cdot(x-x_0)$

$\sf y-3=\dfrac{5}{2}\cdot(x-2)$

$\sf 2y-6=5x-10$

$\sf 5x-2y-10+6=0$

$\sf 5x-2y-4=0$