Question 146253


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



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



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



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



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=2/3}}} and the coordinates of the given point *[Tex \LARGE \left\(0,2\right\)].



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



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



{{{y-2=(-3/2)x+(-3/2)(-0)}}} Distribute



{{{y-2=(-3/2)x+0}}} Multiply



{{{y=(-3/2)x+0+2}}} Add 2 to both sides. 



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



So the equation of the line perpendicular to {{{2x-3y=6}}} that goes through the point *[Tex \LARGE \left\(0,2\right\)] is {{{y=(-3/2)x+2}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(2/3)x-2,(-3/2)x+2)
circle(0,2,0.08),
circle(0,2,0.10),
circle(0,2,0.12))}}}Graph of the original equation {{{y=(2/3)x-2}}} (red) and the perpendicular line {{{y=(-3/2)x+2}}} (green) through the point *[Tex \LARGE \left\(0,2\right\)].