Question

5)De a forma geral da equação da reta r em cada caso. ​

Answer 1

Resposta:

Explicação passo a passo:

A)

(0,4)

(5,0)

0 4 1 0 4

5 0 1 5 0

x y 1 x y

0.0.1 +4.1.x + 1.5.y - [1.0.x +0.1.y+4.5.1]= 0

0 + 4x + 5y - [0+0+20] = 0

4x + 5y - 20 = 0

__________________

B)

(-2,0)

(1, -4)

-2 0 1 -2 0

1 -4 1 1 -4

x y 1 x y

(-2).(-4).1 + 0.1.x + 1.1.y - [1.(-4).x - 2.1.y + 0.1.1] = 0

8 + 0 + y - [- 4x - 2y + 0] = 0

8 + y + 4x + 2y = 0

4x + 3y + 8 = 0

Answer 2

$\Large\boxed{\begin{array}{l}\tt a)~\sf Vamos~obter~a~equac_{\!\!,}\tilde ao~segment\acute aria\\\sf da~reta~e~passar~para~a~forma~geral.\\\sf A(0,4)~~B(5,0)\\\sf\dfrac{x}{5}+\dfrac{y}{4}=1\\\sf multiplicando~por~20~temos:\\\sf \diagdown\!\!\!\!\!\!20\cdot\dfrac{x}{\diagdown\!\!\!5}+\diagdown\!\!\!\!\!\!20\cdot\dfrac{y}{\diagdown\!\!\!\!4}=1\cdot20\\\sf 4x+5y=20\\\sf passando~para~a~forma~a~geral~temos:\\\sf 4x+5y-20=0\end{array}}$

$\Large\boxed{\begin{array}{l}\tt b)~\sf A~reta~passa~pelos~pontos~A (-2,0)~B(1,-4)\\\underline{\rm c\acute alculo~do~coeficiente~angular:}\\\sf m=\dfrac{y_B-y_A}{x_B-x_A}\\\\\sf m=\dfrac{-4-0}{1-(-2)}\\\sf m=-\dfrac{4}{1+2}=-\dfrac{4}{3}\\\sf adotando~o~ponto~A~temos:\\\sf y=0+\bigg(-\dfrac{4}{3}\bigg)\cdot(x-(-2))\\\sf y=-\dfrac{4}{3}\cdot(x+2)\\\sf y=-\dfrac{4}{3}x-\dfrac{8}{3}\\\sf multiplicando~por~3~temos:\\\sf 3\cdot y=3\cdot-\dfrac{4}{3}x-3\cdot\dfrac{8}{3}\\\sf 3y=-4x-8\end{array}}$$\Large\boxed{\begin{array}{l}\sf Passando~para~a~forma~geral~temos:\\\sf4x+3y+8=0\end{array}}$