Question 184192


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)] and *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(0,-3\right)].



{{{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=-1(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{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)]