Question 189713
{{{4x+3y=5}}} Start with the first equation.



{{{3y=5-4x}}} Subtract {{{4x}}} from both sides.



{{{3y=-4x+5}}} Rearrange the terms.



{{{y=(-4x+5)/(3)}}} Divide both sides by {{{3}}} to isolate y.



{{{y=((-4)/(3))x+(5)/(3)}}} Break up the fraction.



{{{y=-(4/3)x+5/3}}} Reduce.



So we can see that the equation {{{y=-(4/3)x+5/3}}} has a slope {{{m=-4/3}}} and a y-intercept {{{b=5/3}}}.



-------------------------------



{{{3x-4y=4}}} Now move onto the second equation.



{{{-4y=4-3x}}} Subtract {{{3x}}} from both sides.



{{{-4y=-3x+4}}} Rearrange the terms.



{{{y=(-3x+4)/(-4)}}} Divide both sides by {{{-4}}} to isolate y.



{{{y=((-3)/(-4))x+(4)/(-4)}}} Break up the fraction.



{{{y=(3/4)x-1}}} Reduce.



So we can see that the equation {{{y=(3/4)x-1}}} has a slope {{{m=3/4}}} and a y-intercept {{{b=-1}}}.



--------------------------------------



So the slope of the first line is {{{m=-4/3}}} and the slope of the second line is {{{m=3/4}}}.



Notice how the slope of the second line {{{m=3/4}}} is simply the negative reciprocal of the slope of the first line {{{m=-4/3}}}.



In other words, if you flip the fraction of the second slope and change its sign, you'll get the first slope. So this means that {{{y=-(4/3)x+5/3}}} and {{{y=(3/4)x-1}}} are perpendicular lines.



So consequently, this also means that {{{4x+3y=5}}} and {{{3x-4y=4}}} are perpendicular lines.