Question

Find the equation of a line that passes though the point (4, -3) and is perpendicular to the line 6x - 3y= 4. ​

Answer 1

bearing in mind that perpendicular lines have negative reciprocal slopes, hmmm what's the slope of the equation above anyway?

`\bf 6x-3y=4\implies -3y=-6x+4\implies y=\cfrac{-6x+4}{-3} \\\\\\ y=\cfrac{-6x}{-3}+\cfrac{4}{-3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{2}x-\cfrac{4}{3}\qquad \impliedby \begin{array}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}`

`\bf ~\dotfill\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{2\implies \cfrac{2}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{2}}}`

so, we're really looking for the equation of a line whose slope is -1/2 and runs through (4,-3).

`\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{-3})~\hspace{10em} \stackrel{slope}{m}\implies -\cfrac{1}{2} \\\\\\ \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}{(-3)}=\stackrel{m}{-\cfrac{1}{2}}(x-\stackrel{x_1}{4}) \\\\\\ y+3=-\cfrac{1}{2}x+2\implies y=-\cfrac{1}{2}x-1`