Answer:
[tex]\large\boxed{y=-\dfrac{1}{5}x+\dfrac{3}{5}}[/tex]
Step-by-step explanation:
[tex]\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-\text{slope}\\b-\text{y-intercept}\\\\\text{Let}\\\\k:y=m_1x+b_1\\l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\k\ ||\ l\iff m_1=m_2\\=========================[/tex]
[tex]\text{We have}\ y=5x+2\to m_1=5\\\\\text{therefore}\ m_2=-\dfrac{1}{5}.\\\\\text{Put the value of a slope and the coordinates of the given point}\ (-2,\ 1)\\\text{to the equation of a line}\ y=mx+b:\\\\1=-\dfrac{1}{5}(-2)+b\\\\1=\dfrac{2}{5}+b\qquad\text{subtract}\ \dfrac{2}{5}\ \text{from both isdes}\\\\1-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{2}{5}+b\\\\\dfrac{5}{5}-\dfrac{2}{5}=b\to b=\dfrac{3}{5}\\\\\text{Finally:}\\\\y=-\dfrac{1}{5}x+\dfrac{3}{5}[/tex]
