Question 207553


{{{3x+2y=0}}} Start with the given equation.



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



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



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



We can see that the equation {{{y=-(3/2)x}}} has a slope {{{m=-3/2}}} and a y-intercept {{{b=0}}}.



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\(-2,-5\right\)].



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



{{{y--5=(2/3)(x--2)}}} Plug in {{{m=2/3}}}, {{{x[1]=-2}}}, and {{{y[1]=-5}}}



{{{y--5=(2/3)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y+5=(2/3)(x+2)}}} Rewrite {{{y--5}}} as {{{y+5}}}



{{{y+5=(2/3)x+(2/3)(2)}}} Distribute



{{{y+5=(2/3)x+4/3}}} Multiply



{{{y=(2/3)x+4/3-5}}} Subtract 5 from both sides. 



{{{y=(2/3)x-11/3}}} Combine like terms. note: If you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>.



So the equation of the line perpendicular to {{{3x+2y=0}}} that goes through the point *[Tex \LARGE \left\(-2,-5\right\)] is {{{y=(2/3)x-11/3}}}.



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,(2/3)x-11/3)
circle(-2,-5,0.08),
circle(-2,-5,0.10),
circle(-2,-5,0.12))}}}Graph of the original equation {{{y=-(3/2)x}}} (red) and the perpendicular line {{{y=(2/3)x-11/3}}} (green) through the point *[Tex \LARGE \left\(-2,-5\right\)].