determine a equação geral da reta que passa pelo ponto P(1;-3) é perpendicular a reta r:2x-y+2=0
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Bom dia Juliana!
Solução!
Vamos chamar a reta perpendicular de s.
Vamos determinar o coeficiente angular da reta r.
![r:2x-y+2=0\\\\\
y=2x+2\\\\\\
\boxed{mr=2}\Rightarrow Coeficiente~~\^Angular.\\\\\\
mr.ms=-1\\\\\\ 2.ms=-1\\\\\
ms= -\dfrac{1}{2}\\\\\\
ms=- \dfrac{1}{12}\Rightarrow Coeficiente~~\^Angular~~da~~reta~~perpendicular!](https://tex.z-dn.net/?f=r%3A2x-y%2B2%3D0%5C%5C%5C%5C%5C%0Ay%3D2x%2B2%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bmr%3D2%7D%5CRightarrow+Coeficiente%7E%7E%5C%5EAngular.%5C%5C%5C%5C%5C%5C%0Amr.ms%3D-1%5C%5C%5C%5C%5C%5C+2.ms%3D-1%5C%5C%5C%5C%5C%0Ams%3D+-%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%5C%5C+%0Ams%3D-+%5Cdfrac%7B1%7D%7B12%7D%5CRightarrow+Coeficiente%7E%7E%5C%5EAngular%7E%7Eda%7E%7Ereta%7E%7Eperpendicular%21 "r:2x-y+2=0\\\\\
y=2x+2\\\\\\
\boxed{mr=2}\Rightarrow Coeficiente~~\^Angular.\\\\\\
mr.ms=-1\\\\\\ 2.ms=-1\\\\\
ms= -\dfrac{1}{2}\\\\\\
ms=- \dfrac{1}{12}\Rightarrow Coeficiente~~\^Angular~~da~~reta~~perpendicular!")
![Equac\~ao~~da~~reta.\\\\\\
y-yP=ms(x-xP)\\\\\\\
P(1,-3)\\\\\\
ms=- \dfrac{1}{2}\\\\\\
y-(-3)= -\dfrac{1}{2}(x-1)\\\\\\\
y+3= \dfrac{-x+1}{2}\\\\\\
2y+6=\dfrac{-x+1}{2}\\\\\\
2y+6=-x+1\\\\\\
2y=-x+1-6\\\\\\
y= \dfrac{-x-5}{2} \\\\\\\\\\\\
\boxed{Resposta:Reta~~y= \dfrac{-x-5}{2}}](https://tex.z-dn.net/?f=Equac%5C%7Eao%7E%7Eda%7E%7Ereta.%5C%5C%5C%5C%5C%5C%0Ay-yP%3Dms%28x-xP%29%5C%5C%5C%5C%5C%5C%5C%0AP%281%2C-3%29%5C%5C%5C%5C%5C%5C%0Ams%3D-+%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0Ay-%28-3%29%3D+-%5Cdfrac%7B1%7D%7B2%7D%28x-1%29%5C%5C%5C%5C%5C%5C%5C%0Ay%2B3%3D+%5Cdfrac%7B-x%2B1%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0A2y%2B6%3D%5Cdfrac%7B-x%2B1%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0A2y%2B6%3D-x%2B1%5C%5C%5C%5C%5C%5C%0A2y%3D-x%2B1-6%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B-x-5%7D%7B2%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BResposta%3AReta%7E%7Ey%3D+%5Cdfrac%7B-x-5%7D%7B2%7D%7D "Equac\~ao~~da~~reta.\\\\\\
y-yP=ms(x-xP)\\\\\\\
P(1,-3)\\\\\\
ms=- \dfrac{1}{2}\\\\\\
y-(-3)= -\dfrac{1}{2}(x-1)\\\\\\\
y+3= \dfrac{-x+1}{2}\\\\\\
2y+6=\dfrac{-x+1}{2}\\\\\\
2y+6=-x+1\\\\\\
2y=-x+1-6\\\\\\
y= \dfrac{-x-5}{2} \\\\\\\\\\\\
\boxed{Resposta:Reta~~y= \dfrac{-x-5}{2}}")
Bom dia!
Bons estudos!
Solução!
Vamos chamar a reta perpendicular de s.
Vamos determinar o coeficiente angular da reta r.
Bom dia!
Bons estudos!
Respondido por
0
A equação geral da reta pedida tem a forma x+2y+c=0 e como ela passa por
P(1;-3) temos 1+2·(-3)+c=0 ⇒1-6+c=0 ⇒ c=5 logo a equação da reta é
x+2y+5=0
Obs. a reta perpendicular a ax +by +c=0 tem a forma bx-ay+c'=0
P(1;-3) temos 1+2·(-3)+c=0 ⇒1-6+c=0 ⇒ c=5 logo a equação da reta é
x+2y+5=0
Obs. a reta perpendicular a ax +by +c=0 tem a forma bx-ay+c'=0
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