Question

Find the equation of a line perpendicular to - 3x + y = - 5 that contains the point (1, 4) Write the equation in slope intercept form

Answer 1

`y=-(1)/(3)x+(13)/(3)`

Explanation:

Convert the given equation to slope-intercept form

`-3x+y=-5\\y=-5+3x\\y=3x-5`

Find the opposite reciprocal of the slope

`3\rightarrow(1)/(3)\rightarrow-(1)/(3)`

Find the new y-intercept of the equation given the point

`y=-(1)/(3)x+b\\ \\4=-(1)/(3)(1)+b\\ \\4=-(1)/(3)+b\\\\(12)/(3)=-(1)/(3)+b\\ \\ b=(13)/(3)`

Final Equation

`y=-(1)/(3)x+(13)/(3)`

Answer 2

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

`\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}}}`

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

`(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad \qquad \stackrel{slope}{m}\implies -\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}{1})\implies y-4=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y = -\cfrac{1}{3}x+\cfrac{1}{3}+4\implies y = -\cfrac{1}{3}x+\cfrac{13}{3}`