Question 338093


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



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(3,0\right)]. So this means that {{{x[1]=3}}} and {{{y[1]=0}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(0,-3\right)].  So this means that {{{x[2]=0}}} and {{{y[2]=-3}}}.



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



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



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



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



{{{m=1}}} Reduce



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



Now let's use the point slope formula:



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



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



{{{y-0=1x+1(-3)}}} Distribute



{{{y-0=1x-3}}} Multiply



{{{y=1x-3+0}}} Add 0 to both sides. 



{{{y=1x-3}}} Combine like terms. 



{{{y=x-3}}} Simplify



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



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

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

 

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