Question 171404

First let's find the slope of the line through the points *[Tex \LARGE \left(2,5\right)] and *[Tex \LARGE \left(-1,11\right)]



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



{{{m=(11-5)/(-1-2)}}} Plug in {{{y[2]=11}}}, {{{y[1]=5}}}, {{{x[2]=-1}}}, and {{{x[1]=2}}}



{{{m=(6)/(-1-2)}}} Subtract {{{5}}} from {{{11}}} to get {{{6}}}



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



{{{m=-2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(2,5\right)] and *[Tex \LARGE \left(-1,11\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-5=-2(x-2)}}} Plug in {{{m=-2}}}, {{{x[1]=2}}}, and {{{y[1]=5}}}



{{{y-5=-2x+-2(-2)}}} Distribute



{{{y-5=-2x+4}}} Multiply



{{{y=-2x+4+5}}} Add 5 to both sides. 



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



{{{y=-2x+9}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(2,5\right)] and *[Tex \LARGE \left(-1,11\right)] is {{{y=-2x+9}}}



 Notice how the graph of {{{y=-2x+9}}} goes through the points *[Tex \LARGE \left(2,5\right)] and *[Tex \LARGE \left(-1,11\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -5, 15,
 graph( 500, 500, -10, 10, -5, 15,-2x+9),
 circle(2,5,0.08),
 circle(2,5,0.10),
 circle(2,5,0.12),
 circle(-1,11,0.08),
 circle(-1,11,0.10),
 circle(-1,11,0.12)
 )}}} Graph of {{{y=-2x+9}}} through the points *[Tex \LARGE \left(2,5\right)] and *[Tex \LARGE \left(-1,11\right)]