Question 307182

First let's find the slope of the line through the points *[Tex \LARGE \left(-2,-6\right)] and *[Tex \LARGE \left(-1,2\right)]



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

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



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



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



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



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



{{{m=8}}} Reduce



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



Now let's use the point slope formula:



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



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



{{{y--6=8(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y+6=8(x+2)}}} Rewrite {{{y--6}}} as {{{y+6}}}



{{{y+6=8x+8(2)}}} Distribute



{{{y+6=8x+16}}} Multiply



{{{y=8x+16-6}}} Subtract 6 from both sides. 



{{{y=8x+10}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(-2,-6\right)] and *[Tex \LARGE \left(-1,2\right)] is {{{y=8x+10}}}



 Notice how the graph of {{{y=8x+10}}} goes through the points *[Tex \LARGE \left(-2,-6\right)] and *[Tex \LARGE \left(-1,2\right)]. So this visually verifies our answer.

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