Find the equation of the line L which is perpendicular to the line 3x - 2y + 6 = 0 and passes through the point (3, -5)

Let [tex]m_{1}[/tex] and [tex]m_{2}[/tex] denote the slope of the given line and the slope of line [tex]L[/tex], respectively. Two lines in a plane are perpendicular to one another if and only if the product of their slopes is [tex](-1)[/tex]. Therefore, for these two lines to be perpendicular to one another, [tex]m_{1} \, m_{2} = (-1)[/tex].

Since [tex]m_{1} = (3/2)[/tex] according to the equation of the given line, the slope of line [tex]L[/tex] would be:

[tex]\begin{aligned} m_{2} &= \frac{(-1)}{m_{1}} \\ &= \frac{(-1)}{(3/2)} \\ &= \frac{(-2)}{3}\end{aligned}[/tex].

If a line in a plane has slope [tex]m[/tex] and goes through the point [tex](x_{0},\, y_{0})[/tex], the slope-point equation of that line would be [tex](y - y_{0}) = m\, (x - x_{0})[/tex].

Since the line [tex]L[/tex] goes through [tex](3,\, -5)[/tex], the equation of this line in slope-point form would be:

[tex]\displaystyle y - (-5) = \frac{(-2)}{3}\, (x - 3)[/tex].

Rearrange to find the equation of line [tex]L[/tex] in slope-intercept form:

[tex]\displaystyle y = -\frac{2}{3}\, x - 3[/tex].

jdoe0001 jdoe0001 · Answer 2 · 2022-01-16T22:12:53+01:00

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

[tex]3x-2y+6=0\implies 3x-2y=-6\implies -2y=-3x-6 \implies y=\cfrac{-3x-6}{-2} \\\\\\ y=\cfrac{-3x}{-2}-\cfrac{6}{-2}\implies y=\stackrel{\stackrel{\stackrel{m}{\downarrow }}{}}{\cfrac{3}{2}} x+3\qquad \impliedby \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}[/tex]

therefore then

[tex]\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{3}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{3}}}[/tex]

so we're really looking for the equation of a line whose slope is -2/3 and passes through (3 , -5)

[tex](\stackrel{x_1}{3}~,~\stackrel{y_1}{-5})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{2}{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}{(-5)}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{3}) \\\\\\ y+5=-\cfrac{2}{3}x+2\implies y=-\cfrac{2}{3}x-3[/tex]