• Matéria: Matemática
  • Autor: danyelavitoria
  • Perguntado 4 anos atrás

ME AJUDEM POR FAVOR PESSOAL
(UFG-GO) As equações das retas r, s e t são 2x + 3y − 1 = 0, x + y + 1 = 0 e x − 2y + 1 = 0
respectivamente. A reta perpendicular a t, que passa pelo ponto de intersecção das retas r e s,tem por equação:
A) 2x + y - 2 = 0\\B) 2x + y + 11 = 0\\C) 2x + y + 5 = 0\\D) X - 2y + 10 = 0\\E) X - 2y - 11 = 0

Respostas

respondido por: CyberKirito
0

\boxed{\begin{array}{l}\rm (UFG-GO)~As~equac_{\!\!,}\tilde oes~das~retas~r,s~e~t~s\tilde ao~2x+3y-1=0,\\\rm x+y+1=0~e~x-2y+1=0~respectivamente. A~reta\\\rm perpendicular~a~t,que~passa~pelo~ponto~de~intersecc_{\!\!,}\tilde ao~das~retas~r~e~s,\\\rm tem~por~equac_{\!\!,}\tilde ao:\\\tt A)~\sf 2x+y-2=0\\\tt B)~\sf 2x+y+11=0\\\tt C)~\sf 2x+y+5=0\\\tt D)~\sf x-2y+10=0\\\tt E)~\sf x-2y-11=0\end{array}}

\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao:}\\\underline{\rm c\acute alculo~do~ponto~de~encontro~das~retas:}\\\sf 2x+3y-1=0\implies 2x+3y=1\\\sf x+y+1=0\implies x+y=-1\\\begin{cases}\sf 2x+3y=1\\\sf x+y=-1\end{cases}\\\underline{\rm multiplicando~a~2^a~equac_{\!\!,}\tilde ao~por\!~-2~temos:}\\+\underline{\begin{cases}\sf \diagdown\!\!\!\!\!\!2x+3y=1\\\sf-\diagdown\!\!\!\!\!\!2x-2y=2\end{cases}}\\\sf y=3\\\underline{\rm substituindo~na~2^a~equac_{\!\!,}\tilde ao}\\\sf x+y=-1\end{array}}

\boxed{\begin{array}{l}\sf x+3=-1\\\sf x=-3-1\\\sf x=-4\\\underline{\rm assim,o~ponto~de~intersecc_{\!\!,}\tilde ao~\acute e}~\sf P(-4,3)\\\sf x-2y+1=0\\\sf 2y=x+1\\\sf y=\dfrac{1}{2}x+\dfrac{1}{2}\\\sf o~coeficiente~angular~da~reta~t~\acute e~m_t=\dfrac{1}{2}\\\sf para~que~duas~retas~sejam~perpendiculares\\\sf \acute e~preciso~que~o~produto~dos~coeficientes~angulares\\\sf seja~igual~a~-1.\\\sf representando~por~u~a~reta~perpendicular~a~t\\\sf podemos~escrever: m_t\cdot m_u=-1\end{array}}

\boxed{\begin{array}{l}\sf\dfrac{1}{2}\cdot m_u=-1\\\sf m_u=2\cdot(-1)\\\sf m_u=-2\\\underline{\rm a~equac_{\!\!,}\tilde ao~pedida~\acute e}\\\sf y=y_0+m_u(x-x_0)\\\sf y=3-2\cdot(x-(-4))\\\sf y=3-2\cdot(x+4)\\\sf y=3-2x-8\\\sf y=-2x-5\\\underline{\rm escrevendo~a~equac_{\!\!,}\tilde ao~na~forma~geral}\\\sf 2x+y+5=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~C}}}}\end{array}}

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