Question 527243


First let's find the slope of the line through the points *[Tex \LARGE \left(-1,-2\right)] and *[Tex \LARGE \left(6,5\right)]



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

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(6,5\right)].  So this means that {{{x[2]=6}}} and {{{y[2]=5}}}.



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



{{{m=(5--2)/(6--1)}}} Plug in {{{y[2]=5}}}, {{{y[1]=-2}}}, {{{x[2]=6}}}, and {{{x[1]=-1}}}



{{{m=(7)/(6--1)}}} Subtract {{{-2}}} from {{{5}}} to get {{{7}}}



{{{m=(7)/(7)}}} Subtract {{{-1}}} from {{{6}}} to get {{{7}}}



{{{m=1}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-1,-2\right)] and *[Tex \LARGE \left(6,5\right)] is {{{m=1}}}



Now let's use the point slope formula:



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



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



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



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



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



{{{y+2=1x+1}}} Multiply



{{{y=1x+1-2}}} Subtract 2 from both sides. 



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



{{{y=x-1}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(-1,-2\right)] and *[Tex \LARGE \left(6,5\right)] is {{{y=x-1}}}



 Notice how the graph of {{{y=x-1}}} goes through the points *[Tex \LARGE \left(-1,-2\right)] and *[Tex \LARGE \left(6,5\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,x-1),
 circle(-1,-2,0.08),
 circle(-1,-2,0.10),
 circle(-1,-2,0.12),
 circle(6,5,0.08),
 circle(6,5,0.10),
 circle(6,5,0.12)
 )}}} Graph of {{{y=x-1}}} through the points *[Tex \LARGE \left(-1,-2\right)] and *[Tex \LARGE \left(6,5\right)]

 



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