267,206 views
5 votes
5 votes
Write the equation of a line perpendicular to 6x-7y=-4 that passes through the point (-6,5).

User Veshraj Joshi
by
2.5k points

2 Answers

9 votes
9 votes

Answer:

Please read and try to understand it :>

Explanation:


6x-7y=-4 is not the standard format for a straight line so you have to rearrange it:


y=(6)/(7) x +(4)/(7)

To find the equation of a perpendicular line to the given equation, you have to first find the negative reciprocal of the gradient:


-(7)/(6)

Since you want to find the equation of a perpendicular line that passes a certain point, you have to substitute the coordinates to the equation to find the y-intercept:


5=-(7)/(6) (-6)+c

Solving:


5=-(7)/(6) (-6) +c\\ \\ 5=(42)/(6) +c\\ \\-(42)/(6) +5=c\\ \\-7+5=c\\\\ -2=c

Final equation:
y=-(7)/(6) x -2

User Marc Talbot
by
2.8k points
26 votes
26 votes

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


6x-7y=-4\implies 6x=7y-4\implies 6x+4=7y \\\\\\ \cfrac{6x+4}{7}=y\implies \qquad \stackrel{\stackrel{m}{\downarrow }}{\cfrac{6}{7}} x+\cfrac{4}{7}=y\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}

now, since we know that slope, let's check for a perpendicular line to it


\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{6}{7}} ~\hfill \stackrel{reciprocal}{\cfrac{7}{6}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{7}{6}}}

so we're really looking for the equation of aline whose slope is -7/6 and that it passes through (-6 , 5)


(\stackrel{x_1}{-6}~,~\stackrel{y_1}{5})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{7}{6} \\\\\\ \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}{5}=\stackrel{m}{- \cfrac{7}{6}}(x-\stackrel{x_1}{(-6)}) \implies y -5= -\cfrac{7}{6} (x +6) \\\\\\ y-5=-\cfrac{7}{6}x-7\implies {\LARGE \begin{array}{llll} y=-\cfrac{7}{6}x-2 \end{array}}

User Xneg
by
2.6k points