Question

as retas (r) 2x+7y=3 e (s) 3x-2y=-8 se cortam num ponto P.
Achar a equação da reta perpendicular a r pelo ponto P.

Answer 1

Inicialmente, encontraremos o ponto P calculando a interseção entre as retas r e s:

$r:~2x+7y=3\iff 7y=-2x+3\iff y=-\frac{2}{7}x+\frac{3}{7}\\\\ s:~3x-2y=-8\iff-2y=-3x-8\iff y=\frac{3}{2}x+4\\\\\\\\ r\cap s=P\Longrightarrow -\frac{2}{7}x+\frac{3}{7}=\frac{3}{2}x+4\iff -2x+3=\frac{21}{2}x+28\iff\\\\-4x+6=21x+56\iff 25x=-50\iff \boxed{x_p=-2}\\\\ \\\\ y=\frac{3}{2}x+4\Longrightarrow y_p=\frac{3}{2}x_p+4\iff y_p=\frac{3}{2}\cdot(-2)+4\iff\\\\y_p=-3+4\iff \boxed{y_p=1}$

Logo, o ponto P é o ponto $P(-2,1).$ Agora, queremos encontrar o coeficiente angular da reta perpendicular a r. Seja p a reta perpendicular. Devido a isso, podemos dizer:

$m_p\cdot m_r=-1\Longrightarrow m_p\cdot\left(-\dfrac{2}{7}\right)=-1\iff m_p=\dfrac{7}{2}$

Com isso, podemos encontrar a equação da reta p:

$p:~y-y_p=m_p(x-x_p)\Longrightarrow y-(1)=\dfrac{7}{2}(x-(-2))\iff\\\\ y-1=\dfrac{7}{2}(x+2)\iff y-1=\frac{7}{2}x+7\iff y=\frac{7}{2}x+8\iff \\\\\boxed{\boxed{p:~7x-2y+16=0}}$