How to write an equation of the line through the point (-2,1) that is perpendicular to the line 5x+9y=-9

Answer:[tex]\large\boxed{y=\dfrac{5}{9}x+\dfrac{19}{9}}[/tex]Step-by-step explanation:[tex]\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\===================================\\\\\text{We have the equation i the standard form.}\\\text{ Convert it to the slope-intercept form}\ y=mx+b:\\\\5x+9y=-9\qquad\text{subtract}\ 5x\ \text{from both sides}\\\\9y=-5x-9\qquad\text{divide both sides by 9}\\\\y=-\dfrac{5}{9}x-1\to m_1=-\dfrac{5}{9}\\\\m_2=-\dfrac{1}{m_1}\to m_2=-\dfrac{1}{-\frac{5}{9}}=\dfrac{9}{5}[/tex][tex]\text{We have the equation:}\\\\y=\dfrac{5}{9}x+b\\\\\text{Put the coordinates of the point (-2, 1) to the equation:}\\\\1=\dfrac{5}{9}(-2)+b\\\\1=-\dfrac{10}{9}+b\qquad\text{add}\ \dfrac{10}{5}\ \text{to both sides}\\\\\dfrac{19}{9}=b\to b=\dfrac{19}{9}[/tex]