Question 268904

First let's find the slope of the line through the points *[Tex \LARGE \left(-3,7\right)] and *[Tex \LARGE \left(-1,-5\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,7\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=7}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-1,-5\right)].  So this means that {{{x[2]=-1}}} and {{{y[2]=-5}}}.



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(-5-7)/(-1--3)}}} Plug in {{{y[2]=-5}}}, {{{y[1]=7}}}, {{{x[2]=-1}}}, and {{{x[1]=-3}}}



{{{m=(-12)/(-1--3)}}} Subtract {{{7}}} from {{{-5}}} to get {{{-12}}}



{{{m=(-12)/(2)}}} Subtract {{{-3}}} from {{{-1}}} to get {{{2}}}



{{{m=-6}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,7\right)] and *[Tex \LARGE \left(-1,-5\right)] is {{{m=-6}}}



Now let's use the point slope formula:



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



{{{y-7=-6(x--3)}}} Plug in {{{m=-6}}}, {{{x[1]=-3}}}, and {{{y[1]=7}}}



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



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



{{{y-7=-6x-18}}} Multiply



{{{y=-6x-18+7}}} Add 7 to both sides. 



{{{y=-6x-11}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-3,7\right)] and *[Tex \LARGE \left(-1,-5\right)] is {{{y=-6x-11}}}



 Notice how the graph of {{{y=-6x-11}}} goes through the points *[Tex \LARGE \left(-3,7\right)] and *[Tex \LARGE \left(-1,-5\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-6x-11),
 circle(-3,7,0.08),
 circle(-3,7,0.10),
 circle(-3,7,0.12),
 circle(-1,-5,0.08),
 circle(-1,-5,0.10),
 circle(-1,-5,0.12)
 )}}} Graph of {{{y=-6x-11}}} through the points *[Tex \LARGE \left(-3,7\right)] and *[Tex \LARGE \left(-1,-5\right)]