Question 180760


{{{2x+3y=2}}} Start with the first equation.



{{{3y=-2x+2}}} Rearrange the terms.



{{{y=(-2x+2)/(3)}}} Divide both sides by {{{3}}} to isolate y.



{{{y=((-2)/(3))x+(2)/(3)}}} Break up the fraction.



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



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


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{{{3x-2y=4}}} Now move onto the second equation.



{{{-2y=-3x+4}}} Rearrange the terms.



{{{y=(-3x+4)/(-2)}}} Divide both sides by {{{-2}}} to isolate y.



{{{y=((-3)/(-2))x+(4)/(-2)}}} Break up the fraction.



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



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



So the slope of the first line is {{{m=-2/3}}} and the slope of the second line is {{{m=3/2}}}.



Notice how the slope of the second line {{{m=3/2}}} is simply the negative reciprocal of the slope of the first line {{{m=-2/3}}}.



In other words, if you flip the fraction of the second slope and change its sign, you'll get the first slope. 



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Answer:



So this means that {{{2x+3y=2}}} and {{{3x-2y=4}}} are perpendicular lines.