Question 612948


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



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,5\right)]. So this means that {{{x[1]=-3}}} and {{{y[1]=5}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(1,-3\right)].  So this means that {{{x[2]=1}}} and {{{y[2]=-3}}}.



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



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



{{{m=(-8)/(1--3)}}} Subtract {{{5}}} from {{{-3}}} to get {{{-8}}}



{{{m=(-8)/(4)}}} Subtract {{{-3}}} from {{{1}}} to get {{{4}}}



{{{m=-2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,5\right)] and *[Tex \LARGE \left(1,-3\right)] is {{{m=-2}}}



Now let's use the point slope formula:



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



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



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



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



{{{y-5=-2x-6}}} Multiply



{{{y=-2x-6+5}}} Add 5 to both sides. 



{{{y=-2x-1}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-3,5\right)] and *[Tex \LARGE \left(1,-3\right)] is {{{y=-2x-1}}}



 Notice how the graph of {{{y=-2x-1}}} goes through the points *[Tex \LARGE \left(-3,5\right)] and *[Tex \LARGE \left(1,-3\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-2x-1),
 circle(-3,5,0.08),
 circle(-3,5,0.10),
 circle(-3,5,0.12),
 circle(1,-3,0.08),
 circle(1,-3,0.10),
 circle(1,-3,0.12)
 )}}} Graph of {{{y=-2x-1}}} through the points *[Tex \LARGE \left(-3,5\right)] and *[Tex \LARGE \left(1,-3\right)]