Question

Qual é o coeficiente angular e linear de cada reta?
a) A (1,1) B (0,2)

b) A (1, - 2) B (2, - 5)

c) A (2,4) B (0,3)

d) A (- 2,5) B (4, - 3)​

Answer 1

Resposta:

a) Coef angular -1   Coef Linear 2

b) Coef angular -3   Coef Linear 1

c) Coef angular $\frac{1}{2}$   Coef Linear 3

d) Coef angular $- \frac{4}{3}$   Coef Linear $-\frac{7}{3}$  

Explicação passo-a-passo:

a) $m =\frac{y_{2} -y_{1} }{x_{2} - x_{1} } =$ $m =\frac{2 -1 }{0 - 1 } = \frac{1}{-1} = -1$ coeficiente angular -1

eq. reduzida da reta

$y - y_{1} = m( x - x_{1}) => y -1 = -1(x -1) => y -1 = -x +1 => y = -x + 2$

coeficiente linear 2

b)  $m =\frac{y_{2} -y_{1} }{x_{2} - x_{1} } =$ $m =\frac{- 5 -(-2) }{2 - 1 } = \frac{-5+2}{1} =\frac{-3}{1} = -3$ coeficiente angular -3

eq. reduzida da reta

$y - y_{1} = m( x - x_{1}) => y -(-2) = -3(x -1) => y +2 = -3x +3 => y = -3x + 1$

coeficiente linear 1

c)  $m =\frac{y_{2} -y_{1} }{x_{2} - x_{1} } =$ $m =\frac{3 - 4 }{0 - 2 } = \frac{-1}{-2} =\frac{1}{2}$

eq. reduzida da reta

$y - y_{1} = m( x - x_{1}) => y -4 = \frac{1}{2} (x -2) => y -4 = \frac{1}{2} x -\frac{2}{2} =>\\ y -4 = \frac{1}{2} x - 1 => y = \frac{1}{2} x - 1 (+4) = y = \frac{1}{2} x + 3$

coeficiente linear 3

d) $m =\frac{y_{2} -y_{1} }{x_{2} - x_{1} } =$ $m =\frac{-3 - 5 }{4 -(-2) } = \frac{-8}{4+2} =\frac{-8}{6} => -\frac{4}{3}$

eq. reduzida da reta

$y - y_{1} = m( x - x_{1}) => y -5 = -\frac{4}{3} (x -(-2) => y -5 = -\frac{4}{3}( x +2 ) =>\\ \\y -5 = - \frac{4}{3} x - \frac{8}{3} => y = - \frac{4}{3} x - (\frac{8}{3} + 5) =>\\ y = \frac{4}{3} x - (\frac{8}{3} +\frac{15}{3} ) => y = \frac{4}{3} x - (+\frac{7}{3} ) => y = \frac{4}{3} x - \frac{7}{3}$

coeficiente linear $-\frac{7}{3}$