Question

1) A reta r determina um angulo de 120º com reta s,cujo coeficiente angular é - 1/3. O coeficiente angular de r é ?

2)As retas r: 3x - y + 5 = 0 e s: kx + 2y - 7 = 0 são concorrentes. Então quanto é o valor de k?

Answer 1

Explicação passo-a-passo:

1)

$\sf tg~\alpha=\dfrac{m_r-m_s}{1+m_r\cdot m_s}$

$\sf tg~120^{\circ}=\dfrac{m_r-\Big(-\frac{1}{3}\Big)}{1+m_r\cdot\Big(-\frac{1}{3}\Big)}$

$\sf -\sqrt{3}=\dfrac{m_r+\frac{1}{3}}{1-\frac{m_r}{3}}$

$\sf -\sqrt{3}=\dfrac{\frac{3m_r+1}{3}}{\frac{3-m_r}{3}}$

$\sf \dfrac{\frac{3m_r+1}{3}}{\frac{3-m_r}{3}}=-\sqrt{3}$

$\sf \dfrac{3m_r+1}{3}\cdot\dfrac{3}{3-m_r}=-\sqrt{3}$

$\sf \dfrac{3m_r+1}{3-m_r}=-\sqrt{3}$

$\sf 3m_r+1=(3-m_r)\cdot(-\sqrt{3})$

$\sf 3m_r+1=-3\sqrt{3}+m_r\cdot\sqrt{3}$

$\sf 3m_r-m_r\cdot\sqrt{3}=-3\sqrt{3}-1$

$\sf m_r\cdot(3-\sqrt{3})=-3\sqrt{3}-1$

$\sf m_r=\dfrac{-3\sqrt{3}-1}{3-\sqrt{3}}$

$\sf m_r=\dfrac{-3\sqrt{3}-1}{3-\sqrt{3}}\cdot\dfrac{3+\sqrt{3}}{3+\sqrt{3}}$

$\sf m_r=\dfrac{-9\sqrt{3}-3-3\cdot3-\sqrt{3}}{3^2-(\sqrt{3})^2}$

$\sf m_r=\dfrac{-9\sqrt{3}-3-9+\sqrt{3}}{9-3}$

$\sf m_r=\dfrac{-8\sqrt{3}-12}{6}$

$\sf m_r=\dfrac{-4\sqrt{3}-6}{3}$

$\sf \red{m_r=\dfrac{-2\cdot(2\sqrt{3}+3)}{3}}$

2)

=> reta r

$\sf 3x-y+5=0$

$\sf y=3x-5$

$\sf m_r=3$

=> reta s

$\sf kx+2y-7=0$

$\sf 2y=-kx+7$

$\sf y=\dfrac{-k}{2}+\dfrac{7}{2}$

$\sf m_s=\dfrac{-k}{2}$

Se duas retas são concorrentes, então os seus coeficientes angulares são diferentes

$\sf -\dfrac{k}{2} \ne 3$

$\sf -k \ne 2\cdot3$

$\sf -k \ne 6~~~~\cdot(-1)$

$\sf \red{k \ne -6}$