Question 198754

{{{-3y=1-2x}}} Subtract {{{2x}}} from both sides.

{{{-3y=-2x+1}}} Rearrange the terms.

{{{y=(-2x+1)/(-3)}}} Divide both sides by {{{-3}}} to isolate y.

{{{y=((-2)/(-3))x+(1)/(-3)}}} Break up the fraction.

{{{y=(2/3)x-1/3}}} Reduce.

We can see that the equation {{{y=(2/3)x-1/3}}} has a slope {{{m=2/3}}} and a y-intercept {{{b=-1/3}}}.

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=2/3}}} and the coordinates of the given point *[Tex \LARGE \left$$-4,8\right$$].

{{{y-8=(-3/2)(x--4)}}} Plug in {{{m=-3/2}}}, {{{x[1]=-4}}}, and {{{y[1]=8}}}

{{{y-8=(-3/2)(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}

{{{y-8=(-3/2)x+(-3/2)(4)}}} Distribute

{{{y-8=(-3/2)x-6}}} Multiply

{{{y=(-3/2)x-6+8}}} Add 8 to both sides.

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

So the equation of the line perpendicular to {{{2x-3y=1}}} that goes through the point *[Tex \LARGE \left$$-4,8\right$$] is {{{y=(-3/2)x+2}}}.

The equation in function notation would then be {{{f(x)=(-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-1/3,(-3/2)x+2)
circle(-4,8,0.08),
circle(-4,8,0.10),
circle(-4,8,0.12))}}}Graph of the original equation {{{y=(2/3)x-1/3}}} (red) and the perpendicular line {{{y=(-3/2)x+2}}} (green) through the point *[Tex \LARGE \left$$-4,8\right$$].