Question 158073

{{{3x-8y=-18}}} Start with the first equation.



{{{-8y=-18-3x}}} Subtract {{{3x}}} from both sides.



{{{-8y=-3x-18}}} Rearrange the terms.



{{{y=(-3x-18)/(-8)}}} Divide both sides by {{{-8}}} to isolate y.



{{{y=((-3)/(-8))x+(-18)/(-8)}}} Break up the fraction.



{{{y=(3/8)x+9/4}}} Reduce.



So we can see that the equation {{{y=(3/8)x+9/4}}} has a slope {{{m=3/8}}} and a y-intercept {{{b=9/4}}}.



{{{32x+12y=-18}}} Now move onto the second equation



{{{12y=-18-32x}}} Subtract {{{32x}}} from both sides.



{{{12y=-32x-18}}} Rearrange the terms.



{{{y=(-32x-18)/(12)}}} Divide both sides by {{{12}}} to isolate y.



{{{y=((-32)/(12))x+(-18)/(12)}}} Break up the fraction.



{{{y=-(8/3)x-3/2}}} Reduce.



So we can see that the equation {{{y=-(8/3)x-3/2}}} has a slope {{{m=-8/3}}} and a y-intercept {{{b=-3/2}}}.



So the slope of the first line is {{{m=3/8}}} and the slope of the second line is {{{m=-8/3}}}.



Notice how the slope of the second line {{{m=-8/3}}} is simply the negative reciprocal of the slope of the first line {{{m=3/8}}}.



In other words, if you flip the fraction of the second slope and change its sign, you'll get the first slope (or vice versa). So this means that {{{y=(3/8)x+9/4}}} and {{{y=-(8/3)x-3/2}}} are perpendicular lines.