Question 397155


{{{y+3x=8}}} Start with the given equation.



{{{y=8-3x}}} Subtract {{{3x}}} from both sides.



{{{y=-3x+8}}} Rearrange the terms.



We can see that the equation {{{y=-3*x+8}}} has a slope {{{m=-3}}} and a y-intercept {{{b=8}}}.



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\(2,3\right\)].



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



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



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



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



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



{{{y=(1/3)x+7/3}}} 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+y=8}}} that goes through the point *[Tex \LARGE \left\(2,3\right\)] is {{{y=(1/3)x+7/3}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-3*x+8,(1/3)x+7/3)
circle(2,3,0.08),
circle(2,3,0.10),
circle(2,3,0.12))}}}


Graph of the original equation {{{y=-3*x+8}}} (red) and the perpendicular line {{{y=(1/3)x+7/3}}} (green) through the point *[Tex \LARGE \left\(2,3\right\)]. 



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Jim