Question 174179

{{{2y=3x-6}}} Start with the given equation.



{{{y=(3x-6)/(2)}}} Divide both sides by {{{2}}} to isolate y.



{{{y=((3)/(2))x+(-6)/(2)}}} Break up the fraction.



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



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



Now to find the slope of the perpendicular line, simply flip the slope {{{m=3/2}}} to get {{{m=2/3}}}. Now change the sign to get {{{m=-2/3}}}. So the perpendicular slope is {{{m=-2/3}}}.



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=3/2}}} and the coordinates of the given point *[Tex \LARGE \left\(3,3\right\)].



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



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



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



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



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



{{{y=(-2/3)x+5}}} Combine like terms. 



So the equation of the line perpendicular to {{{-3x+2y=-6}}} that goes through the point *[Tex \LARGE \left\(3,3\right\)] is {{{y=(-2/3)x+5}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(3/2)x-3,(-2/3)x+5)
circle(3,3,0.08),
circle(3,3,0.10),
circle(3,3,0.12))}}}Graph of the original equation {{{y=(3/2)x-3}}} (red) and the perpendicular line {{{y=(-2/3)x+5}}} (green) through the point *[Tex \LARGE \left\(3,3\right\)].