Question 644375


First let's find the slope of the line through the points *[Tex \LARGE \left(-6,-2\right)] and *[Tex \LARGE \left(-8,-3\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-6,-2\right)]. So this means that {{{x[1]=-6}}} and {{{y[1]=-2}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-8,-3\right)].  So this means that {{{x[2]=-8}}} and {{{y[2]=-3}}}.



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



{{{m=(-3--2)/(-8--6)}}} Plug in {{{y[2]=-3}}}, {{{y[1]=-2}}}, {{{x[2]=-8}}}, and {{{x[1]=-6}}}



{{{m=(-1)/(-8--6)}}} Subtract {{{-2}}} from {{{-3}}} to get {{{-1}}}



{{{m=(-1)/(-2)}}} Subtract {{{-6}}} from {{{-8}}} to get {{{-2}}}



{{{m=1/2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-6,-2\right)] and *[Tex \LARGE \left(-8,-3\right)] is {{{m=1/2}}}



Now let's use the point slope formula:



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



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



{{{y--2=(1/2)(x+6)}}} Rewrite {{{x--6}}} as {{{x+6}}}



{{{y+2=(1/2)(x+6)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(1/2)x+(1/2)(6)}}} Distribute



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



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



{{{y=(1/2)x+1}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-6,-2\right)] and *[Tex \LARGE \left(-8,-3\right)] is {{{y=(1/2)x+1}}}



 Notice how the graph of {{{y=(1/2)x+1}}} goes through the points *[Tex \LARGE \left(-6,-2\right)] and *[Tex \LARGE \left(-8,-3\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(1/2)x+1),
 circle(-6,-2,0.08),
 circle(-6,-2,0.10),
 circle(-6,-2,0.12),
 circle(-8,-3,0.08),
 circle(-8,-3,0.10),
 circle(-8,-3,0.12)
 )}}} Graph of {{{y=(1/2)x+1}}} through the points *[Tex \LARGE \left(-6,-2\right)] and *[Tex \LARGE \left(-8,-3\right)]