• Matéria: Matemática
  • Autor: jonatasandrade1977
  • Perguntado 6 anos atrás

02. Verifique se as retas r e s abaixo são perpendiculares em cada um dos casos:

a) r: x + 7y - 10 = 0 e s: y = 7x + 3

b) r: x - y + 7 = 0 e s: 2x + 5y - 7 = 0

03. As retas 3x + 2y – 1 = 0 e -4x + 6y – 10 = 0 são paralelas? Justifique sua resposta.

04. Prove que as retas s: x + 2y – 1 = 0 e r: 4x – 2y +12 = 0 são perpendiculares.

05. Encontre a equação da reta s, perpendicular à reta t: 2x + 3y – 4 =0, sabendo que ela passa pelo ponto

P(3,4).

Respostas

respondido por: CyberKirito
10

Posição relativa de duas retas:

Considere duas retas

\mathtt{r_{1}:y=m_{1}x+n_{1}}\\\mathtt{r_{2}:y=m_{2}x+n_{2}}

\huge\mathsf{Ent\tilde{a}o:}

\mathtt{m_{1}=m_{2}~e~n_{1}\ne~n_{2}\to~r_{1}//r_{2}}\\\mathtt{m_{1}=m_{2}~e~n_{1}=n_{2}\to~r_{1}\,e\,r_{2}\,concidem}\\\mathtt{m_{1}\ne~m_{2}\to~r_{1}\,X\,r_{2}}}\\\mathtt{m_{1}\times\,m_{2}=-1\to~r_{1}\perp~r_{2}}

02.

\mathtt{a)~r:x+7y-10=0~e~s:~y=7x+3}\\\mathtt{m_{r}=-\dfrac{1}{7}~~m_{s}=7}\\\mathtt{m_{r}.m_{s}=-\dfrac{1}{7}.7=-1\to~r_{1}\perp~r_{2}}

\mathtt{b)~r:x-y+7=0~e~s:~2x+5y-7=0}\\\mathtt{m_{r}=1~m_{s}=-\dfrac{2}{5}}\\\mathtt{m_{r}.m_{s}=1.-\dfrac{2}{5}=-\dfrac{2}{5}\ne~-1\to~r\not\perp~s}

3)

\mathtt{r:3x+2y-1=0~e~s:-4x+6y-10=0}\\\mathtt{m_{r}=-\dfrac{3}{2}~m_{s}=\dfrac{2}{3}}\\\mathtt{m_{r}.m_{s}=-\dfrac{3}{2}.\dfrac{2}{3}=-1\to~r\perp~s}

4)

\mathtt{s:x+2y-1=0~e~r:4x-2y+12=0}\\\mathtt{m_{s}=-\dfrac{1}{2}~m_{r}=2}\\\mathtt{m_{s}.m_{r}=-\dfrac{1}{2}.2=-1\to~s\perp~r}

5)

\mathtt{Se\, s\, \'e\\, perpendicular\, a\, t\, ent\tilde{a}o}\\\mathtt{m_{s}.m_{t}=-1}.

\mathtt{t:~2x+3y-4=0\to~m_{t}=-\dfrac{2}{3}}\\\mathtt{m_{t}.m_{s}=-1}\\\mathtt{-\dfrac{2}{3}.m_{s}=-1\to~m_{s}=\dfrac{3}{2}}

\mathtt{A~reta~t~passa~pelos~pontos~A(0,\dfrac{4}{3})~e~B(2,0)}

\mathtt{Adotando~o~ponto~A(0,\dfrac{4}{3})~temos:}\\\huge\boxed{\boxed{\mathtt{s:~y=\dfrac{4}{3}+\dfrac{3}{2}x}}}}

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