Question 190110

First let's find the slope of the line through the points *[Tex \LARGE \left(1,5\right)] and *[Tex \LARGE \left(3,6\right)]



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



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



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



{{{m=(1)/(3-1)}}} Subtract {{{5}}} from {{{6}}} to get {{{1}}}



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



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



{{{y-5=(1/2)x+(1/2)(-1)}}} Distribute



{{{y-5=(1/2)x-1/2}}} Multiply



{{{y=(1/2)x-1/2+5}}} Add 5 to both sides. 



{{{y=(1/2)x+9/2}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(1,5\right)] and *[Tex \LARGE \left(3,6\right)] is {{{y=(1/2)x+9/2}}}



 Notice how the graph of {{{y=(1/2)x+9/2}}} goes through the points *[Tex \LARGE \left(1,5\right)] and *[Tex \LARGE \left(3,6\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(1/2)x+9/2),
 circle(1,5,0.08),
 circle(1,5,0.10),
 circle(1,5,0.12),
 circle(3,6,0.08),
 circle(3,6,0.10),
 circle(3,6,0.12)
 )}}} Graph of {{{y=(1/2)x+9/2}}} through the points *[Tex \LARGE \left(1,5\right)] and *[Tex \LARGE \left(3,6\right)]