Question 207628

First let's find the slope of the line through the points *[Tex \LARGE \left(6,-3\right)] and *[Tex \LARGE \left(4,1\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(6,-3\right)]. So this means that {{{x[1]=6}}} and {{{y[1]=-3}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(4,1\right)].  So this means that {{{x[2]=4}}} and {{{y[2]=1}}}.



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



{{{m=(1--3)/(4-6)}}} Plug in {{{y[2]=1}}}, {{{y[1]=-3}}}, {{{x[2]=4}}}, and {{{x[1]=6}}}



{{{m=(4)/(4-6)}}} Subtract {{{-3}}} from {{{1}}} to get {{{4}}}



{{{m=(4)/(-2)}}} Subtract {{{6}}} from {{{4}}} to get {{{-2}}}



{{{m=-2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(6,-3\right)] and *[Tex \LARGE \left(4,1\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--3=-2(x-6)}}} Plug in {{{m=-2}}}, {{{x[1]=6}}}, and {{{y[1]=-3}}}



{{{y+3=-2(x-6)}}} Rewrite {{{y--3}}} as {{{y+3}}}



{{{y+3=-2x+-2(-6)}}} Distribute



{{{y+3=-2x+12}}} Multiply



{{{y=-2x+12-3}}} Subtract 3 from both sides. 



{{{y=-2x+9}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(6,-3\right)] and *[Tex \LARGE \left(4,1\right)] is {{{y=-2x+9}}}



 Notice how the graph of {{{y=-2x+9}}} goes through the points *[Tex \LARGE \left(6,-3\right)] and *[Tex \LARGE \left(4,1\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-2x+9),
 circle(6,-3,0.08),
 circle(6,-3,0.10),
 circle(6,-3,0.12),
 circle(4,1,0.08),
 circle(4,1,0.10),
 circle(4,1,0.12)
 )}}} Graph of {{{y=-2x+9}}} through the points *[Tex \LARGE \left(6,-3\right)] and *[Tex \LARGE \left(4,1\right)]