Matemática, perguntado por c00973757, 9 meses atrás

1- as retas r: 3x+y-3=0 e s: 6x+2y-6=0 são; 2- as retas r: x+y-2=0 e s: x+y-4=0

Anexos:

Soluções para a tarefa

Respondido por CyberKirito
1

Posição relativa de duas retas

Considere duas retas

\mathsf{\ell_{1}:m_{1}x+n_{1}}\\\mathsf{\ell_{2}:m_{2}x+n_{2}}.

Então

\mathsf{m_{1}=m_{2}~e~n_{1}\ne n_{2}\implies\,\ell_{1}~e~\ell_{2}~paralelas}\\\mathsf{m_{1}=m_{2}~e~n_{1}=n_{2}\implies\,\ell_{1}~e~\ell_{2}~coincidem}\\\mathsf{m_{1}\ne m_{2}\implies\,\ell_{1}~e~\ell_{2}~concorrentes}\\\mathsf{m_{1}\cdot m_{2}=-1\implies\,\ell_{1}~e~\ell_{2}~perpendiculares}

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1)

\mathsf{r:~3x+y-3=0\implies~y=-3x+3~m_{1}=-3~~n_{1}=3}\\\mathsf{s:6x+2y-6=0\implies~y=-3x+3~m_{2}=-3~~n_{2}=3}\\\mathsf{r~e~s~concidem}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{\maltese~~alternativa~~c}}}}}

2)

\mathsf{r:x+y-2=0\implies~y=-x+2~m_{1}=-1~~n_{1}=2}\\\mathsf{s:y-4=0\implies\,y=0x+4~m_{2}=0~~n_{2}=4}\\\mathsf{r~e~s~concorrentes}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{\maltese~~alternativa~~a}}}}}

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