Question 148201

First let's find the slope of the line through the points *[Tex \LARGE \left(-2,3\right)] and *[Tex \LARGE \left(1,5\right)]



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



{{{m=(5-3)/(1--2)}}} Plug in {{{y[2]=5}}}, {{{y[1]=3}}}, {{{x[2]=1}}}, {{{x[1]=-2}}}, , 



{{{m=(2)/(1--2)}}} Subtract {{{3}}} from {{{5}}} to get {{{2}}}



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



So the slope of the line that goes through the points *[Tex \LARGE \left(-2,3\right)] and *[Tex \LARGE \left(1,5\right)] is {{{m=2/3}}}



Now let's use the point slope formula:



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



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



{{{y-3=(2/3)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y-3=(2/3)x+(2/3)(2)}}} Distribute



{{{y-3=(2/3)x+4/3}}} Multiply



{{{y=(2/3)x+4/3+3}}} Add 3 to both sides. 



{{{y=(2/3)x+4/3+9/3}}} Rewrite {{{3}}} as {{{9/3}}}. 



{{{y=(2/3)x+13/3}}} Combine the fractions.




So the equation that goes through the points *[Tex \LARGE \left(-2,3\right)] and *[Tex \LARGE \left(1,5\right)] is {{{y=(2/3)x+13/3}}}



 Notice how the graph of {{{y=(2/3)x+13/3}}} goes through the points *[Tex \LARGE \left(-2,3\right)] and *[Tex \LARGE \left(1,5\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(2/3)x+13/3),
 circle(-2,3,0.08),
 circle(-2,3,0.10),
 circle(-2,3,0.12),
 circle(1,5,0.08),
 circle(1,5,0.10),
 circle(1,5,0.12)
 )}}} Graph of {{{y=(2/3)x+13/3}}} through the points *[Tex \LARGE \left(-2,3\right)] and *[Tex \LARGE \left(1,5\right)]