Question 347816
We can see that the equation {{{y=3x+5}}} has a slope {{{m=3}}} and a y-intercept {{{b=5}}}.



Now to find the slope of the perpendicular line, simply flip the slope {{{m=3}}} to get {{{m=1/3}}}. Now change the sign to get {{{m=-1/3}}}. So the perpendicular slope is {{{m=-1/3}}}.



Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=-1/3}}} and the coordinates of the given point *[Tex \LARGE \left\(-1,2\right\)].



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



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



{{{y-2=(-1/3)(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y-2=(-1/3)x+(-1/3)(1)}}} Distribute



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



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



{{{y=(-1/3)x+5/3}}} 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 {{{y=3x+5}}} that goes through the point *[Tex \LARGE \left\(-1,2\right\)] is {{{y=(-1/3)x+5/3}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,3x+5,(-1/3)x+5/3)
circle(-1,2,0.08),
circle(-1,2,0.10),
circle(-1,2,0.12))}}}


Graph of the original equation {{{y=3x+5}}} (red) and the perpendicular line {{{y=(-1/3)x+5/3}}} (green) through the point *[Tex \LARGE \left\(-1,2\right\)]. 



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Jim