Question

A reta r e s representadas no plano cartesiano a seguir são perpendiculares. Determine a inclinação e o coeficiente angular da reta s

(Tô estudando pela prova passada e até hoje não sei como resolver isso)
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Answer 1

$\large\boxed{\begin{array}{l}\rm a\,inclinac_{\!\!,}\tilde ao\,da\,reta\,s\\\rm \acute e\,um\,\hat angulo\,externo\,ao\,tri\hat angulo\\\rm chamando\,de\,\beta\,o\,\hat angulo\,dado\\\rm vamos\,usar\,o\,teorema\,do\,\hat angulo\,externo\\\rm que\,diz\,o\,\hat angulo\,externo\,\acute e\,igual\,a\,soma\\\rm dos\,\hat angulos\,internos\,n\tilde ao\,adjacentes.\\\rm \beta=90^\circ+30^\circ=120^\circ\\\rm ou\,seja\,a\,inclinac_{\!\!,}\tilde ao\,de\,s\,\acute e\,120^\circ\\\rm \end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf o\,coeficiente\,angular\,de\,s\,ser\acute a}\\\rm m_s=tg(\beta)\\\rm m_s=tg(120^\circ)\\\rm m_s=-\sqrt{3}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf outra\,forma\,de\,resolver:}\\\rm a\,inclinac_{\!\!,}\tilde ao\,de\,r\,\acute e\,30^\circ\\\rm e\,o\,coeficiente\,angular\\\rm m_r=tg(30^\circ)=\dfrac{\sqrt{3}}{3}\\\\\rm s\perp r\iff m_s\cdot m_r=-1\\\rm m_s\cdot\dfrac{\sqrt{3}}{3}=-1\\\rm m_s=-\dfrac{3}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}\\\\\rm m_s=-\dfrac{\backslash\!\!\!3\sqrt{3}}{\backslash\!\!\!3}\\\\\rm m_s=-\sqrt{3}\end{array}}$

$\large\boxed{\begin{array}{l}\rm chamando\,de\,\alpha\,a\,inclinac_{\!\!,}\tilde ao\,de\,s\,temos:\\\rm m_s=tg(\alpha)\\\rm tg(\alpha)=-\sqrt{3}\\\rm \alpha=arc\,tg(-\sqrt{3})\\\rm \alpha=120^\circ\end{array}}$