Question

(UECE) Seja r a reta que passa pelos pontos P1(–2, 1) e P2(5, 3). Se r intercepta os eixos coordenadas nos pontos M(m, 0) e N(0, n), então o valor de 14/11.(n – m) é: a) 6 b) 7 c) 8 d) 9 Gabarito: D
Resolução: coeficiente angular da reta = (3 - 1)/[5 - (-2)] = 2/7
equação da reta passando por P2(5, 3)---> y - 3 = (2/7).(x - 5) ----> y = (2/7).x + 11/7
M(m, 0) ---> 0 = (2/7).m + 11/7 ---> m = - 11/2 N(0, n) ---> n = (2/7).0 + 11/7 ---> n = 11/7 (14/11).(11/7 + 11/2) = (14/11).(22 + 77)/14 = 99/11 = 9
DE ONDE SAIU ESSE 11/7 eu não consigo achar já quebrei a cabeça aqui ...

Answer 1

Explicação passo-a-passo:

• Coeficiente angular

$\sf m=\dfrac{y_{P_{2}}-y_{P_{1}}}{x_{P_{2}}-x_{P_{1}}}$

$\sf m=\dfrac{3-1}{5-(-2)}$

$\sf m=\dfrac{3-1}{5+2}$

$\sf m=\dfrac{2}{7}$

• Equação da reta

$\sf y-y_0=m\cdot(x-x_0)$

$\sf y-3=\dfrac{2}{7}\cdot(x-5)$

$\sf y-3=\dfrac{2x}{7}-\dfrac{10}{7}$

$\sf y=\dfrac{2x}{7}-\dfrac{10}{7}+3$

$\sf y=\dfrac{2x}{7}-\dfrac{10}{7}+\dfrac{3\cdot7}{7}$

$\sf y=\dfrac{2x}{7}-\dfrac{10}{7}+\dfrac{21}{7}$

$\sf y=\dfrac{2x}{7}+\dfrac{11}{7}$

A reta intercepta o eixo x quando y = 0:

$\sf \dfrac{2x}{7}+\dfrac{11}{7}=0$

$\sf 2x+11=0$

$\sf 2x=-11$

$\sf x=\dfrac{-11}{2}~\rightarrow~m=\dfrac{-11}{2}$

A reta intercepta o eixo y quando x = 0:

$\sf y=\dfrac{2\cdot0}{7}+\dfrac{11}{7}$

$\sf y=\dfrac{11}{7}~\rightarrow~n=\dfrac{11}{7}$

Logo:

$\sf \dfrac{14}{11}\cdot(n-m)$

$\sf =\dfrac{14}{11}\cdot\left(\dfrac{11}{7}+\dfrac{11}{2}\right)$

$\sf =\dfrac{14}{11}\cdot\left(\dfrac{2\cdot11+7\cdot11}{14}\right)$

$\sf =\dfrac{14}{11}\cdot\left(\dfrac{22+77}{14}\right)$

$\sf =\dfrac{14}{11}\cdot\dfrac{99}{14}$

$\sf =\dfrac{99}{11}$

$\sf =9$

Letra D