Question

A equação da reta perpendicular ao segmento de reta de extremos A(-2,5) e B(6,-1) e que passa pelo ponto médio desse segmento é

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Equac_{\!\!,}\tilde ao~da~reta}\\\underline{\rm que~passa~por~A(x_0,y_0)}\\\underline{\rm e~possui~coeficiente~angular~m}\\\huge\boxed{\boxed{\boxed{\boxed{\sf y=y_0+m\cdot (x-x_0)}}}}\\\\\underline{\rm Coeficiente~angular~da~reta}\\\underline{\rm que~passa~por~A(x_A,y_A)~e~B(x_B,y_B)}\\\huge\boxed{\boxed{\boxed{\boxed{\sf m=\dfrac{y_B-y_A}{x_B-x_A}}}}}\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Ponto~m\acute edio~de~um~segmento.}\\\rm Dado~os~pontos~A(x_A,y_A)~e~B(x_B,y_B)\\\rm o~ponto~m\acute edio~que~passa~por~A~e~B\\\rm \acute e~representado~por~M(x_M,y_M)\\\rm e~dado~por\\\huge\boxed{\boxed{\boxed{\boxed{\sf x_M=\dfrac{x_A+x_B}{2}}}}}~\rm e~\huge\boxed{\boxed{\boxed{\boxed{\sf y_M=\dfrac{y_A+y_B}{2}}}}}\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Condic_{\!\!,}\tilde ao~de~perpendicularidade}\\\underline{\rm de~duas~retas}\\\rm Dada~as~retas\\\rm ~r:ax+by+c=0~e~s: mx+ny+k=0\\\rm r\perp s\iff m_r\cdot m_s=-1\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm C\acute alculo~do~coeficiente~angular:}\\\\\sf m_r=\dfrac{ y_B-y_A}{x_B-x_A}\\\\\sf m_r=\dfrac{-1-5}{6-(-2)}=\dfrac{-6}{8}=-\dfrac{3}{4}\\\underline{\rm C\acute alculo~do~ponto~m\acute edio:}\\\\\sf x_M=\dfrac{x_A+x_B}{2}=\dfrac{-2+6}{2}=\dfrac{4}{2}=2\\\\\sf y_M=\dfrac{y_A+y_B}{2}=\dfrac{5-1}{2}=\dfrac{4}{2}=2\\\sf M(2,2).\\\sf r\perp s\iff m_r\cdot m_s=-1\\\sf -\dfrac{3}{4}\cdot m_s=-1\implies m_s=\dfrac{4}{3}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf y=2+\dfrac{4}{3}\cdot(x-2)\\\\\sf y=2+\dfrac{4}{3}x-\dfrac{8}{3}\cdot3\\\\\sf 3y=6+4x-8\\\sf s:4x-3y-2=0\end{array}}$