Question 189692


First let's find the slope of the line through the points *[Tex \LARGE \left(2,3\right)] and *[Tex \LARGE \left(6,5\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(2,3\right)] and *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(6,5\right)].



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



{{{m=(5-3)/(6-2)}}} Plug in {{{y[2]=5}}}, {{{y[1]=3}}}, {{{x[2]=6}}}, and {{{x[1]=2}}}



{{{m=(2)/(6-2)}}} Subtract {{{3}}} from {{{5}}} to get {{{2}}}



{{{m=(2)/(4)}}} Subtract {{{2}}} from {{{6}}} to get {{{4}}}



{{{m=1/2}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(2,3\right)] and *[Tex \LARGE \left(6,5\right)] is {{{m=1/2}}}



Now let's use the point slope formula:



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



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



{{{y-3=(1/2)x+(1/2)(-2)}}} Distribute



{{{y-3=(1/2)x-1}}} Multiply



{{{y=(1/2)x-1+3}}} Add 3 to both sides. 



{{{y=(1/2)x+2}}} Combine like terms. 



{{{y=(1/2)x+2}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(2,3\right)] and *[Tex \LARGE \left(6,5\right)] is {{{y=(1/2)x+2}}}



 Notice how the graph of {{{y=(1/2)x+2}}} goes through the points *[Tex \LARGE \left(2,3\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,(1/2)x+2),
 circle(2,3,0.08),
 circle(2,3,0.10),
 circle(2,3,0.12),
 circle(6,5,0.08),
 circle(6,5,0.10),
 circle(6,5,0.12)
 )}}} Graph of {{{y=(1/2)x+2}}} through the points *[Tex \LARGE \left(2,3\right)] and *[Tex \LARGE \left(6,5\right)]