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

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$-3x+y=-5\implies y=\stackrel{\stackrel{m}{\downarrow }}{3} x-5\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

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)

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y-4=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y = -\cfrac{1}{3}x+\cfrac{1}{3}+4\implies y = -\cfrac{1}{3}x+\cfrac{13}{3}$

Answer 2

Answer:

$y=-\frac{1}{3}x+\frac{13}{3}$

Step-by-step 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

$3\rightarrow\frac{1}{3}\rightarrow-\frac{1}{3}$

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

$y=-\frac{1}{3}x+b\\ \\4=-\frac{1}{3}(1)+b\\ \\4=-\frac{1}{3}+b\\\\\frac{12}{3}=-\frac{1}{3}+b\\ \\ b=\frac{13}{3}$

Final Equation

$y=-\frac{1}{3}x+\frac{13}{3}$