Question

Find the equation of the line that contains the point (-1,-10) and is parallel to the line 2x+3y=5 . Write the equation in​ slope-intercept form, if possible.

Answer 1

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

`2x+3y=5\implies 3y=-2x+5\implies y=\cfrac{-2x+5}{3} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+\cfrac{5}{3}\qquad \impliedby \begin{array} \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 line whose slope is -2/3 and that it passes through (-1 , -10)

`(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-10})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{2}{3} \\\\\\ \begin{array} \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}{(-10)}=\stackrel{m}{- \cfrac{2}{3}}(x-\stackrel{x_1}{(-1)}) \implies y +10= -\cfrac{2}{3} (x +1) \\\\\\ y+10=-\cfrac{2}{3}x-\cfrac{2}{3}\implies y=-\cfrac{2}{3}x-\cfrac{2}{3}-10\implies {\Large \begin{array}{llll} y=-\cfrac{2}{3}x-\cfrac{32}{3} \end{array}}`