EQUAÇÃO DA RETA R QUE PASSA POR A(2,3) E E PERPENDICULAR A RETA S:X-Y-1=0
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Boa noite Paulo!
Solução!\\\\\
![A(2,3)\\\\\
S:x-y-1=0](https://tex.z-dn.net/?f=A%282%2C3%29%5C%5C%5C%5C%5C%0AS%3Ax-y-1%3D0 "A(2,3)\\\\\
S:x-y-1=0")
Vamos colocar a reta s na forma reduzida para encontramos o coeficiente angular.
![S:x-y-1=0\\\\\
-y=-x+1\\\\
y=x-1\\\\
m=1](https://tex.z-dn.net/?f=S%3Ax-y-1%3D0%5C%5C%5C%5C%5C%0A-y%3D-x%2B1%5C%5C%5C%5C%0Ay%3Dx-1%5C%5C%5C%5C%0Am%3D1 "S:x-y-1=0\\\\\
-y=-x+1\\\\
y=x-1\\\\
m=1")
Para que as retas sejam perpendiculares,o coeficiente tem que ser o inverso.
![m(s).m(r)=-1\\\\\
1.m(r)=-1\\\\
m(r)= \dfrac{-1}{1}\\\\\
m(r)=-1\\\\\\
Equac\~ao ~~da~~reta~~r\\\\\\\
y-yA=m(r)(x-xA)\\\\\\\
y-3=-1(x-2)\\\\\\
y-3=-x+2\\\\\
y=-x+2+3\\\\\
r:y=-x+5\\\\\\
s:x-y-1\perp r:y=-x+5\\\\\\\\
\boxed{Resposta:y=-x+5~~ou~~x+y-5=0}](https://tex.z-dn.net/?f=m%28s%29.m%28r%29%3D-1%5C%5C%5C%5C%5C%0A1.m%28r%29%3D-1%5C%5C%5C%5C%0Am%28r%29%3D+%5Cdfrac%7B-1%7D%7B1%7D%5C%5C%5C%5C%5C%0Am%28r%29%3D-1%5C%5C%5C%5C%5C%5C%0AEquac%5C%7Eao+%7E%7Eda%7E%7Ereta%7E%7Er%5C%5C%5C%5C%5C%5C%5C%0Ay-yA%3Dm%28r%29%28x-xA%29%5C%5C%5C%5C%5C%5C%5C%0Ay-3%3D-1%28x-2%29%5C%5C%5C%5C%5C%5C%0Ay-3%3D-x%2B2%5C%5C%5C%5C%5C%0Ay%3D-x%2B2%2B3%5C%5C%5C%5C%5C%0Ar%3Ay%3D-x%2B5%5C%5C%5C%5C%5C%5C%0As%3Ax-y-1%5Cperp+r%3Ay%3D-x%2B5%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BResposta%3Ay%3D-x%2B5%7E%7Eou%7E%7Ex%2By-5%3D0%7D%0A+ "m(s).m(r)=-1\\\\\
1.m(r)=-1\\\\
m(r)= \dfrac{-1}{1}\\\\\
m(r)=-1\\\\\\
Equac\~ao ~~da~~reta~~r\\\\\\\
y-yA=m(r)(x-xA)\\\\\\\
y-3=-1(x-2)\\\\\\
y-3=-x+2\\\\\
y=-x+2+3\\\\\
r:y=-x+5\\\\\\
s:x-y-1\perp r:y=-x+5\\\\\\\\
\boxed{Resposta:y=-x+5~~ou~~x+y-5=0}")
Boa noite!
Bons estudos!
Solução!\\\\\
Vamos colocar a reta s na forma reduzida para encontramos o coeficiente angular.
Para que as retas sejam perpendiculares,o coeficiente tem que ser o inverso.
Boa noite!
Bons estudos!
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