Question 335970
Any line that is perpendicular to a line with a slope of {{{3/2}}} will have a slope of {{{-2/3}}} (flip the fraction and change the sign)



So our perpendicular line has a slope of {{{-2/3}}} and goes through the point (3, 0)



Now we can use the point slope formula to find the equation of the line. The point slope formula is:



{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point



{{{y-0=(-2/3)(x-3)}}} Plug in {{{m=-2/3}}}, {{{x[1]=3}}}, and {{{y[1]=0}}} 



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



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



{{{y=(-2/3)x+2}}} Multiply {{{-2/3}}} and {{{-3}}} to get {{{2}}}



So the equation of the line is {{{y=(-2/3)x+2}}}



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Jim