Question 157964


First let's find the slope of the line through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(6,-4\right)]



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



{{{m=(-4-0)/(6-3)}}} Plug in {{{y[2]=-4}}}, {{{y[1]=0}}}, {{{x[2]=6}}}, and {{{x[1]=3}}}



{{{m=(-4)/(6-3)}}} Subtract {{{0}}} from {{{-4}}} to get {{{-4}}}



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



So the slope of the line that goes through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(6,-4\right)] is {{{m=-4/3}}}



Now let's use the point slope formula:



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



{{{y-0=(-4/3)(x-3)}}} Plug in {{{m=-4/3}}}, {{{x[1]=3}}}, and {{{y[1]=0}}}



{{{y-0=(-4/3)x+(-4/3)(-3)}}} Distribute



{{{y-0=(-4/3)x+4}}} Multiply



{{{y=(-4/3)x+4+0}}} Add 0 to both sides. 



{{{y=(-4/3)x+4}}} Combine like terms. 



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



So the equation that goes through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(6,-4\right)] is {{{y=(-4/3)x+4}}}



 Notice how the graph of {{{y=(-4/3)x+4}}} goes through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(6,-4\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(-4/3)x+4),
 circle(3,0,0.08),
 circle(3,0,0.10),
 circle(3,0,0.12),
 circle(6,-4,0.08),
 circle(6,-4,0.10),
 circle(6,-4,0.12)
 )}}} Graph of {{{y=(-4/3)x+4}}} through the points *[Tex \LARGE \left(3,0\right)] and *[Tex \LARGE \left(6,-4\right)]