Question 243011

{{{2x-3y=5}}} Start with the given equation.



{{{-3y=5-2x}}} Subtract {{{2x}}} from both sides.



{{{-3y=-2x+5}}} Rearrange the terms.



{{{y=(-2x+5)/(-3)}}} Divide both sides by {{{-3}}} to isolate y.



{{{y=((-2)/(-3))x+(5)/(-3)}}} Break up the fraction.



{{{y=(2/3)x-5/3}}} Reduce.



We can see that the equation {{{y=(2/3)x-5/3}}} has a slope {{{m=2/3}}} and a y-intercept {{{b=-5/3}}}.



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=-3/2}}} and the coordinates of the given point *[Tex \LARGE \left\(-2,6\right\)].



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



{{{y-6=(-3/2)(x--2)}}} Plug in {{{m=-3/2}}}, {{{x[1]=-2}}}, and {{{y[1]=6}}}



{{{y-6=(-3/2)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y-6=(-3/2)x+(-3/2)(2)}}} Distribute



{{{y-6=(-3/2)x-3}}} Multiply



{{{y=(-3/2)x-3+6}}} Add 6 to both sides. 



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



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



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-5/3,(-3/2)x+3)
circle(-2,6,0.08),
circle(-2,6,0.10),
circle(-2,6,0.12))}}}


Graph of the original equation {{{y=(2/3)x-5/3}}} (red) and the perpendicular line {{{y=(-3/2)x+3}}} (green) through the point *[Tex \LARGE \left\(-2,6\right\)].