Question

What is the slopeOf a line that is perpendicular to a line who’s equation is 8y-5x=11

Answer 1

first off, let's put 8y - 5x = 11, in a slope-intercept form, so we can see what's its slope

$\bf 8y-5x=11\implies 8y=5x+11\implies y=\cfrac{5x+11}{8} \\\\\\ y=\stackrel{slope}{\cfrac{5}{8}}x+\cfrac{11}{8}$

so the slope of this line is 5/8, well then

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{5}{8}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{8}{5}}\qquad \stackrel{negative~reciprocal}{-\cfrac{8}{5}}}$