Question

Equation of the line that is parallel to x-3y=9 and passes through the point (-10,9)

Answer 1

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

`x-3y=9\implies -3y=-x+9\implies y=\cfrac{-x+9}{-3} \\\\\\ y=\cfrac{-x}{-3}+\cfrac{9}{-3}\implies y=\cfrac{1}{3}x-3\qquad \impliedby \begin{array}c \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}`

so we're really looking for the equation of a likne whose slope is 1/3 and it passes through (-10 , 9)

`(\stackrel{x_1}{-10}~,~\stackrel{y_1}{9})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{3} \\\\\\ \begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{9}=\stackrel{m}{ \cfrac{1}{3}}(x-\stackrel{x_1}{(-10)}) \implies y -9= \cfrac{1}{3} (x +10) \\\\\\ y-9=\cfrac{1}{3}x+\cfrac{10}{3}\implies y=\cfrac{1}{3}x+\cfrac{10}{3}+9\implies {\Large \begin{array}{llll} y=\cfrac{1}{3}x+\cfrac{37}{3} \end{array}}`