Question

C-determine a equação da reta paralela à reta r dada e que passe pelo ponto p, conforme as situações seguintes :​

Answer 1

$\large\boxed{\begin{array}{l}\tt a)~\sf P(-5,6)~~r:2x+4y-3=0\implies m_r=-\dfrac{2}{4}=-\dfrac{1}{2}\\\sf r\parallel s\iff m_s=m_r=-\dfrac{1}{2}\\\sf y=6-\dfrac{1}{2}(x-(-5))\\\\\sf y=6-\dfrac{1}{2}(x+5)\\\\\sf y=\dfrac{12}{2}-\dfrac{x}{2}+\dfrac{5}{2}\\\\\sf s: y=\dfrac{17-x}{2}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt b)~\sf P(2,-3)~r:3x-7y+3=0\\\sf m_r=-\dfrac{3}{-7}=\dfrac{3}{7}\\\sf r\parallel s\iff m_s=m_r=\dfrac{3}{7}\\\sf y=-3+\dfrac{3}{7}(x-2)\\\\\sf y=-\dfrac{21}{7}+\dfrac{3}{7}x-\dfrac{6}{7}\\\\\sf s: y=\dfrac{-27+3x}{7}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt c )~\sf P(-2,-3)~~r:4x-5y+1=0\\\sf m_r=-\dfrac{4}{-5}=\dfrac{4}{5}\\\sf r\parallel s\iff m_r=m_s=\dfrac{4}{5}\\\\\sf y=-3+\dfrac{4}{5}(x-[-2])\\\\\sf y=-\dfrac{15}{5}+\dfrac{4}{5}(x+2)\\\\\sf y=-\dfrac{15}{5}+\dfrac{4}{5}x+\dfrac{8}{5}\\\\\sf y=\dfrac{-15+4x+8}{5}\\\\\sf s: y=\dfrac{4x-7}{5}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt d)~\sf P\bigg(-\dfrac{3}{2},-2\bigg)~~r:3x-2y+2=0\\\sf m_r=-\dfrac{3}{-2}=\dfrac{3}{2}\\\sf r\parallel s\iff m_r=m_s=\dfrac{3}{2}\\\\\sf y=-2+\dfrac{3}{2}\bigg(x-\bigg[-\dfrac{3}{2}\bigg]\bigg)\\\\\sf y=-\dfrac{4}{2}+\dfrac{3}{2}\bigg(x+\dfrac{3}{2}\bigg)\\\\\sf y=-\dfrac{4}{2}+\dfrac{3}{2}x+\dfrac{9}{4}\\\\\sf y=\dfrac{-8+6x+9}{4}\\\\\sf s: y=\dfrac{6x+1}{4}\end{array}}$