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Exercício de Intersecção de retas
Exercício:
a)Determine o ponto P de intersecção entre retas de equações 2x-5y+3=0 e x-3y-7=0
b) Determine a equação da reta que é perpendicular à reta de equação 4x+y-1=0 e passa pelo ponto P mencionado á cima.
Informações Adicionais:
Sei que deve-se fazer o sistema para achar o X e o Y
eu consigo achar o Y.
2x-5y+3=0
x-3y-7=0 (x -2)
exclui os X’s, fica -5y +6y +3+14 = 0 // y = 17
mas, e pra achar o X?
nao tem como igualar -5y com -3y para excluir os Y’s ..ou tem?
como?
Cara Hevellyn,
Você tem que fazer um sistema para achar Y
2x-5y = -3
x-3y=+7 (x-2)
2x-5y = -3
-2x+6y = -14
Y = -17
Agora basta você mudar o y em uma das equações. Escolhi a 2ª:
x – 3(-17) =+7
x = -44
P = (-44,-17)
da onde surgiu o (x-2)?
Não surgiu, eu multipliquei toda a segunda equação por -2, para poder cortar com a de cima no sistema ^-^
=*
Mariana…
A) Achando o valor de x e y no sistema de equações:
I- 2*x-5*y=-3
II- x-3*y=7 ,obtemos x=-44 e y=-17, logo o ponto P tem coordenadas (-44,-17).
B) Como o coeficiente angular da reta 4*x+y-1=0 é m=-4, agora podemos determinar a equação da reta perpendicular a esta que passa pelo ponto P, logo seu coeficiente será:
n=1/m=-(1/4) e sua equação:
y-(-17)=-(1/4)*[x-(-44)], reduzindo a expressão vem:
y+17=(-1/4)*(x+44), multiplicando ambos os membros por 4, obtemos:
4*y+68=-x-44, organizando a mesma,
x+4*y+112=0 end.
{x+y=4{2x+y=7
{x+y=4
2x+y=5
1)A equação da reta que passa pelo ponto P(2,5) e é paralelo a reta da equação x-y+2=0 é?
2)A equção da reta que passa pelo ponto (-1,2) e forma com o eixo 0x um angulo de 45° é?
3)O valor de k para que a reta k.x -4.y+2k=0 passe pelo ponto de intersecção das retas 2.x-y+3=0 e x+y-9=0 é?
Se puder ajudar tenho um tema pra amanhã e não consigui resolve as quetões 2 e 3 e na primeira to com uma duvida na parte final!
Sabendo que a reta s é paralela a r onde r tem por equação 3x-10y+30=0 e que s contém origem, concluimos que a equação de S é?
2x-y+6=0 e 2x+3y-6=0
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