A equação da reta que passa pelos pontos (-1,-2) e (5,2)
Resposta com calculo por favor !
Soluções para a tarefa
Respondido por
1
Boa tarde Ligia!
Solução!
Vamos determinar o coeficiente angular da reta com essa formula.
![A(-1,-2)\\\\\
B(5,2)\\\\\\\
m= \dfrac{yB-yA}{xB-xA}\\\\\\\
m= \dfrac{2-(-2)}{5-(-1)}\\\\\\\
m= \dfrac{2+2}{5+1}\\\\\\\
m=\dfrac{4}{6}\\\\\\\
\boxed{m= \frac{2}{3} }](https://tex.z-dn.net/?f=A%28-1%2C-2%29%5C%5C%5C%5C%5C%0AB%285%2C2%29%5C%5C%5C%5C%5C%5C%5C%0Am%3D+%5Cdfrac%7ByB-yA%7D%7BxB-xA%7D%5C%5C%5C%5C%5C%5C%5C%0Am%3D+%5Cdfrac%7B2-%28-2%29%7D%7B5-%28-1%29%7D%5C%5C%5C%5C%5C%5C%5C%0A%0Am%3D+%5Cdfrac%7B2%2B2%7D%7B5%2B1%7D%5C%5C%5C%5C%5C%5C%5C%0Am%3D%5Cdfrac%7B4%7D%7B6%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bm%3D+%5Cfrac%7B2%7D%7B3%7D+%7D "A(-1,-2)\\\\\
B(5,2)\\\\\\\
m= \dfrac{yB-yA}{xB-xA}\\\\\\\
m= \dfrac{2-(-2)}{5-(-1)}\\\\\\\
m= \dfrac{2+2}{5+1}\\\\\\\
m=\dfrac{4}{6}\\\\\\\
\boxed{m= \frac{2}{3} }")
Equação da reta!
![A(-1,-2)\\\\\
m= \dfrac{2}{3} \\\\\\
y-yA=m(x-xA)\\\\\\
y+2= \dfrac{2}{3} (x+1)\\\\\\
y+2= \dfrac{2x}{3}+ \dfrac{2}{3}\\\\\\
3y+6=2x+2\\\\\\
3y=2x+2-6\\\\\\
3y=2x-4\\\\\
\boxed{y= \dfrac{2x-4}{3}}](https://tex.z-dn.net/?f=A%28-1%2C-2%29%5C%5C%5C%5C%5C%0Am%3D+%5Cdfrac%7B2%7D%7B3%7D+%5C%5C%5C%5C%5C%5C%0A%0Ay-yA%3Dm%28x-xA%29%5C%5C%5C%5C%5C%5C%0Ay%2B2%3D+%5Cdfrac%7B2%7D%7B3%7D+%28x%2B1%29%5C%5C%5C%5C%5C%5C%0A%0Ay%2B2%3D+%5Cdfrac%7B2x%7D%7B3%7D%2B+%5Cdfrac%7B2%7D%7B3%7D%5C%5C%5C%5C%5C%5C%0A3y%2B6%3D2x%2B2%5C%5C%5C%5C%5C%5C%0A3y%3D2x%2B2-6%5C%5C%5C%5C%5C%5C%0A3y%3D2x-4%5C%5C%5C%5C%5C%0A%5Cboxed%7By%3D+%5Cdfrac%7B2x-4%7D%7B3%7D%7D%0A%0A "A(-1,-2)\\\\\
m= \dfrac{2}{3} \\\\\\
y-yA=m(x-xA)\\\\\\
y+2= \dfrac{2}{3} (x+1)\\\\\\
y+2= \dfrac{2x}{3}+ \dfrac{2}{3}\\\\\\
3y+6=2x+2\\\\\\
3y=2x+2-6\\\\\\
3y=2x-4\\\\\
\boxed{y= \dfrac{2x-4}{3}}")
Boa tarde!
Bons estudos!
Solução!
Vamos determinar o coeficiente angular da reta com essa formula.
Equação da reta!
Boa tarde!
Bons estudos!
Respondido por
2
Oi Ligia
f(x) = ax + b
f(-1) = -a + b = -2
f(5) = 5a + b = 2
5a - (-a) = 2 - (-2)
5a + a = 2 + 2
6a = 4
a = 4/6 = 2/3
-2/3 + b = -6/3
b = -6/3 + 2/3 =-4/3
f(x) = (2x - 4)/3
equação reduzida
y = (2x - 4)/3
equação geral
2x - 3y - 4 = 0
.
f(x) = ax + b
f(-1) = -a + b = -2
f(5) = 5a + b = 2
5a - (-a) = 2 - (-2)
5a + a = 2 + 2
6a = 4
a = 4/6 = 2/3
-2/3 + b = -6/3
b = -6/3 + 2/3 =-4/3
f(x) = (2x - 4)/3
equação reduzida
y = (2x - 4)/3
equação geral
2x - 3y - 4 = 0
.
Perguntas interessantes
Geografia,
7 meses atrás
Português,
7 meses atrás
Matemática,
1 ano atrás
Sociologia,
1 ano atrás
Ed. Física,
1 ano atrás