Question

1- as retas r: 2x+y-6=0 e s: -2x+y+1=0 são: 2- as retas r: x+y-5=0 e s: -x+y=0 são:

Answer 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)

$\mathsf{r:~2x+y-6=0\implies y=-2x+6~~m_{1}=-2}\\\mathsf{-2x+y+1=0\implies y=2x-1~~m_{2}=2}\\\mathsf{m_{1}\ne m_{2}\implies r~e~s~concorrentes}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{\maltese~~alternativa~~b}}}}}$

2- as retas r: x+y-5=0 e s: -x+y=0 são:

$\mathsf{r:~x+y-5=0\implies~y=-x+5~~m_{1}=-1}\\\mathsf{s:-x+y=0\implies~y=x~~m_{2}=1}\\\mathsf{m_{1}\cdot m_{2}=-1\cdot1=-1}\\\mathsf{r~e~s~~perpendiculares}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{\maltese~~alternativa~~d}}}}}$

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