Question

a reta R (x -7y +13 = 0) forma um ângulo de 45º com a reta S, que passa pela intersecção das retas 3x + 2Y -9 = 0 e 10x + y -13 = 0. ache a reta S

Answer 1

$\displaystyle \text R : \text x-7\text y+13=0 \\\\ \text R : \text y = \frac{\text x}{7}+\frac{13}{7} \to \boxed{\text m_\text R =\frac{1}{7}}$

O reta S passa pela interseção das retas :

$3\text x+2\text y-9=0\\\\ 10\text x+\text y-13=0 \\\\ \text{achando a interse{\c c}{\~a}o}: \\\\ \text y = 13-10\text x \\\\ 3\text x+2(13-10\text x)-9=0 \\\\ 3\text x+26-20\text x-9=0 \\\\ 17\text x=17 \to \boxed{\text x = 1} \\\\ \text y =13-10.1 \to \boxed{\text y =3}$

Então a equação da reta S será do tipo :

$\text y-3=\text m_\text S (\text x-1)$

Reta R com a reta S forma um ângulo de 45º, e sabemos que :

$\displaystyle \text {tg}(\theta)=|\frac{\text m_ \text R-\text m_\text S}{1+\text{m}_\text R.\text m_\text S } |$

ou seja :

$\displaystyle \text{tg}(45^\circ)=|\frac{\displaystyle \frac{1}{7}-\text m_\text S}{\displaystyle 1+\frac{1}{7}.\text m_\text S}| \\\\\\ 1 = |\frac{\displaystyle \frac{1-7\text m_\text S}{7}}{ \displaystyle \frac{7+\text m_\text S}{7}}| \to 1=|\frac{1-7\text m_\text S}{7+\text m_\text S}| \\\\ \underline{\text{Da{\'i} temos dois casos}}: \\\\ 1^\circ \text{caso }: \\\\ 1 = \frac{1-7\text m_\text S }{7+\text m_\text S} \\\\ 7+\text m_\text S =1-7\text m_\text S\\\\ 6=-8\text m_\text S$

$\displaystyle \boxed{\text m_ \text S= \frac{-3}{4}}$

$\displaystyle \underline{2^\circ \text{caso}}: \\\\ -1 =\frac{1-7\text m_\text S }{7+\text m_\text S} \\\\ -7-\text m_\text S = 1-7\text m_\text S \\\\ 6\text m_\text S =8 \\\\ \boxed{\text m_\text S = \frac{4}{3} }$

Equação da Reta S:

$\displaystyle \text S : \text y-3=\frac{-3}{4}(\text x-1) \\\\ \text S : 4\text y-12=-3\text x+3 \\\\ \huge\boxed{\text S : 3\text x+4\text y-15=0\ }\checkmark$