Question 98508
Lets find the equation of the line through (-6,5) and (-3, -2)



First lets find the slope through the points ({{{-6}}},{{{5}}}) and ({{{-3}}},{{{-2}}})


{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{-6}}},{{{5}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{-3}}},{{{-2}}}))


{{{m=(-2-5)/(-3--6)}}} Plug in {{{y[2]=-2}}},{{{y[1]=5}}},{{{x[2]=-3}}},{{{x[1]=-6}}}  (these are the coordinates of given points)


{{{m= -7/3}}} Subtract the terms in the numerator {{{-2-5}}} to get {{{-7}}}.  Subtract the terms in the denominator {{{-3--6}}} to get {{{3}}}

  

So the slope is

{{{m=-7/3}}}


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Now let's use the point-slope formula to find the equation of the line:




------Point-Slope Formula------
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(\textrm{x_{1},y_{1}}\right)] is one of the given points


So lets use the Point-Slope Formula to find the equation of the line


{{{y-5=(-7/3)(x--6)}}} Plug in {{{m=-7/3}}}, {{{x[1]=-6}}}, and {{{y[1]=5}}} (these values are given)



{{{y-5=(-7/3)(x+6)}}} Rewrite {{{x--6}}} as {{{x+6}}}



{{{y-5=(-7/3)x+(-7/3)(6)}}} Distribute {{{-7/3}}}


{{{y-5=(-7/3)x-14}}} Multiply {{{-7/3}}} and {{{6}}} to get {{{-42/3}}}. Now reduce {{{-42/3}}} to get {{{-14}}}


{{{y=(-7/3)x-14+5}}} Add {{{5}}} to  both sides to isolate y


{{{y=(-7/3)x-9}}} Combine like terms {{{-14}}} and {{{5}}} to get {{{-9}}} 

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Answer:



So the equation of the line which goes through the points ({{{-6}}},{{{5}}}) and ({{{-3}}},{{{-2}}})  is:{{{y=(-7/3)x-9}}}




Now let's find the equation of the line perpendicular to {{{y=(-7/3)x-9}}} and goes through (2,-4)



*[invoke equation_parallel_or_perpendicular "perpendicular", "-7/3", -9, 2, -4]