Question

B. Passes through the point (2, -4) and is parallel to 3x + y = 5

c. Passes through the point (2, -4) and is perpendicular to 3x + y = 5

Answer 1

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

`3x+y=5\implies y=\stackrel{\stackrel{m}{\downarrow }}{-3}x+5\qquad \impliedby \qquad \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 line that has a slope oif -3 and it passes through (2 , -4)

`(\stackrel{x_1}{2}~,~\stackrel{y_1}{-4})\hspace{10em} \stackrel{slope}{m} ~=~ - 3 \\\\\\ \begin{array}c \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}{(-4)}=\stackrel{m}{- 3}(x-\stackrel{x_1}{2}) \implies y +4 = - 3 ( x -2) \\\\\\ y+4=-3x+6\implies {\Large \begin{array}{llll} y=-3x+2 \end{array}}`

now, keeping in mind that perpendicular lines have negative reciprocal slopes

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ -3 \implies \cfrac{-3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-3} \implies \cfrac{1}{ 3 }}}`

so for this one, we're looking for the equation of a line whose slope is 1/3 and it passes through (2 , -4)

`(\stackrel{x_1}{2}~,~\stackrel{y_1}{-4})\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}{(-4)}=\stackrel{m}{ \cfrac{1}{3}}(x-\stackrel{x_1}{2}) \implies y +4 = \cfrac{1}{3} ( x -2) \\\\\\ y+4=\cfrac{1}{3}x-\cfrac{2}{3}\implies y=\cfrac{1}{3}x-\cfrac{2}{3}-4\implies {\Large \begin{array}{llll} y=\cfrac{1}{3}x-\cfrac{14}{3} \end{array}}`