Question 197547

First let's find the slope of the line through the points *[Tex \LARGE \left(6,-2\right)] and *[Tex \LARGE \left(3,-4\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(6,-2\right)] and *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(3,-4\right)].



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



{{{m=(-4--2)/(3-6)}}} Plug in {{{y[2]=-4}}}, {{{y[1]=-2}}}, {{{x[2]=3}}}, and {{{x[1]=6}}}



{{{m=(-2)/(3-6)}}} Subtract {{{-2}}} from {{{-4}}} to get {{{-2}}}



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



{{{m=2/3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(6,-2\right)] and *[Tex \LARGE \left(3,-4\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--2=(2/3)(x-6)}}} Plug in {{{m=2/3}}}, {{{x[1]=6}}}, and {{{y[1]=-2}}}



{{{y+2=(2/3)(x-6)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(2/3)x+(2/3)(-6)}}} Distribute



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



{{{y=(2/3)x-4-2}}} Subtract 2 from both sides. 



{{{y=(2/3)x-6}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(6,-2\right)] and *[Tex \LARGE \left(3,-4\right)] is {{{y=(2/3)x-6}}}



 Notice how the graph of {{{y=(2/3)x-6}}} goes through the points *[Tex \LARGE \left(6,-2\right)] and *[Tex \LARGE \left(3,-4\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(2/3)x-6),
 circle(6,-2,0.08),
 circle(6,-2,0.10),
 circle(6,-2,0.12),
 circle(3,-4,0.08),
 circle(3,-4,0.10),
 circle(3,-4,0.12)
 )}}} Graph of {{{y=(2/3)x-6}}} through the points *[Tex \LARGE \left(6,-2\right)] and *[Tex \LARGE \left(3,-4\right)]