Question 145681
Remember, the "origin" is simply the point (0,0)





First let's find the slope through the points *[Tex \LARGE \left(3,2\right)] and *[Tex \LARGE \left(0,0\right)]



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



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



{{{m=(-2)/(0-3)}}} Subtract {{{2}}} from {{{0}}} to get {{{-2}}}



{{{m=(-2)/(-3)}}} Subtract {{{3}}} from {{{0}}} to get {{{-3}}}



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



So the slope of the line that goes through the points *[Tex \LARGE \left(3,2\right)] and *[Tex \LARGE \left(0,0\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-3)}}} Plug in {{{m=2/3}}}, {{{x[1]=3}}}, and {{{y[1]=2}}}



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



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



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



{{{y=(2/3)x+0}}} Combine like terms. 



{{{y=(2/3)x}}} Remove the trailing zero



So the equation that goes through the points *[Tex \LARGE \left(3,2\right)] and *[Tex \LARGE \left(0,0\right)] is {{{y=(2/3)x}}}



 Notice how the graph of {{{y=(2/3)x}}} goes through the points *[Tex \LARGE \left(3,2\right)] and *[Tex \LARGE \left(0,0\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(2/3)x),
 circle(3,2,0.08),
 circle(3,2,0.10),
 circle(3,2,0.12),
 circle(0,0,0.08),
 circle(0,0,0.10),
 circle(0,0,0.12)
 )}}} Graph of {{{y=(2/3)x}}} through the points *[Tex \LARGE \left(3,2\right)] and *[Tex \LARGE \left(0,0\right)]