Question

Write the equation of a line perpendicular to 3x + 4y = 9 that passes through (8, - 4)

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, let's find the slope of 3x + 4y = 9, by simply putting it in slope-intercept form.

$\bf 3x+4y=9\implies 4y=-3x+9\implies y=-\cfrac{3x+9}{4}\implies y=\stackrel{slope}{-\cfrac{3}{4}}x+\cfrac{9}{4} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{4}{3}}\qquad \stackrel{negative~reciprocal}{+\cfrac{4}{3}}\implies \cfrac{4}{3}}$

so we're really looking for the equation of a line whose slope is 4/3 and runs through 8, -4.

$\bf (\stackrel{x_1}{8}~,~\stackrel{y_1}{-4})~\hspace{10em} slope = m\implies \cfrac{4}{3} \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{4}{3}(x-8) \implies y+4=\cfrac{4}{3}x-\cfrac{32}{3} \\\\\\ y=\cfrac{4}{3}x-\cfrac{32}{3}-4\implies y=\cfrac{4}{3}x-\cfrac{44}{3}$