Perpendicular to 3x-6y=-2 containing points (-3,-8)

Answer:y = -2x - 14Step-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}[/tex][tex]\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 the equation of a line in the standard form.}\\\text{Convert it to the slope intercept from:}\\\\3x-6y=-2\qquad\text{subtract }\ 3x\ \text{from both sides}\\\\-6y=-3x-2\qquad\text{divide both sides by (-6)}\\\\y=\dfrac{-3}{-6}x+\dfrac{-2}{-6}\\\\y=\dfrac{1}{2}x+\dfrac{1}{3}\to m_1=\dfrac{1}{2}[/tex][tex]\text{Therefore}\ m_2=-\dfrac{1}{\frac{1}{2}}=-1\cdot\dfrac{2}{1}=-2.\\\\\text{Put the value of a slope and the coordinates of the point (-3, -8)}\\\text{ to the equation of a line:}\\\\-8=-2(-3)+b\\\\-8=6+b\qquad\text{subtract 6 from both sides}\\\\-14=b\to b=-14\\\\\text{Finally:}\\\\y=-2x-14[/tex]