Question 237558


{{{3x-4y=8}}} Start with the first equation.



{{{-4y=8-3x}}} Subtract {{{3x}}} from both sides.



{{{-4y=-3x+8}}} Rearrange the terms.



{{{y=(-3x+8)/(-4)}}} Divide both sides by {{{-4}}} to isolate y.



{{{y=((-3)/(-4))x+(8)/(-4)}}} Break up the fraction.



{{{y=(3/4)x-2}}} Reduce.



So we can see that the equation {{{y=(3/4)x-2}}} has a slope {{{m=3/4}}} and a y-intercept {{{b=-2}}}.



{{{8x+6y=8}}} Now move onto the second equation.



{{{6y=8-8x}}} Subtract {{{8x}}} from both sides.



{{{6y=-8x+8}}} Rearrange the terms.



{{{y=(-8x+8)/(6)}}} Divide both sides by {{{6}}} to isolate y.



{{{y=((-8)/(6))x+(8)/(6)}}} Break up the fraction.



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



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



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



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



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=(3/4)x-2}}} and {{{y=-(4/3)x+4/3}}} are perpendicular lines. Consequently, this means that {{{3x-4y=8}}} and {{{8x+6y=8}}} are also perpendicular lines.