Question

Calcule o valor de k para que as retas dadas por x +3y - 13 = 0  e kx + y = 0 formem um ângulo de 45 graus.  

Answer 1

Olá, Josué.

Reta r: x + 3y - 13 = 0 ⇒ 3y = -x + 13 ⇒ y = -$\frac13$x + $\frac{13}3$ ⇒ $m_r=-\frac13$

Reta s: kx + y = 0 ⇒ y = -kx ⇒ $m_s$ = -k

Ângulo $\theta$ entre as retas r e s:

$tg\,\theta=|\frac{m_r-m_s}{1+m_rm_s}|\Rightarrow\\\\ tg\,45\º=\left|\frac{-\frac13-(-k)}{1+(-\frac13)(-k)}\right|\Rightarrow\\\\ 1=\left|\frac{-\frac13+k}{1+\frac k3}\right|\Rightarrow\\\\ 1=\left|\frac{\frac{-1+3k}3}{\frac{3+k}3}\right|\Rightarrow\\\\ 1=\left|\frac{-1+3k}{3+k}\right|\Rightarrow\\\\ \begin{cases}\frac{-1+3k}{3+k}=1\Rightarrow -1+3k=3+k\Rightarrow 2k=4\Rightarrow \boxed{k=2}\\\text{ou}\\\frac{-1+3k}{3+k}=-1\Rightarrow -1+3k=-3-k\Rightarrow 4k=-2\Rightarrow \boxed{k=-\frac12}\end{cases}$