Question 307211


First let's find the slope of the line through the points *[Tex \LARGE \left(2,-1\right)] and *[Tex \LARGE \left(0,3\right)]



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

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(0,3\right)].  So this means that {{{x[2]=0}}} and {{{y[2]=3}}}.



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



{{{m=(3--1)/(0-2)}}} Plug in {{{y[2]=3}}}, {{{y[1]=-1}}}, {{{x[2]=0}}}, and {{{x[1]=2}}}



{{{m=(4)/(0-2)}}} Subtract {{{-1}}} from {{{3}}} to get {{{4}}}



{{{m=(4)/(-2)}}} Subtract {{{2}}} from {{{0}}} to get {{{-2}}}



{{{m=-2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(2,-1\right)] and *[Tex \LARGE \left(0,3\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--1=-2(x-2)}}} Plug in {{{m=-2}}}, {{{x[1]=2}}}, and {{{y[1]=-1}}}



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



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



{{{y+1=-2x+4}}} Multiply



{{{y=-2x+4-1}}} Subtract 1 from both sides. 



{{{y=-2x+3}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(2,-1\right)] and *[Tex \LARGE \left(0,3\right)] is {{{y=-2x+3}}}



 Notice how the graph of {{{y=-2x+3}}} goes through the points *[Tex \LARGE \left(2,-1\right)] and *[Tex \LARGE \left(0,3\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-2x+3),
 circle(2,-1,0.08),
 circle(2,-1,0.10),
 circle(2,-1,0.12),
 circle(0,3,0.08),
 circle(0,3,0.10),
 circle(0,3,0.12)
 )}}} Graph of {{{y=-2x+3}}} through the points *[Tex \LARGE \left(2,-1\right)] and *[Tex \LARGE \left(0,3\right)]