Question 186940


First let's find the slope of the line through the points *[Tex \LARGE \left(-3,6\right)] and *[Tex \LARGE \left(-1,0\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-3,6\right)] and *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-1,0\right)].



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



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



{{{m=(-6)/(-1--3)}}} Subtract {{{6}}} from {{{0}}} to get {{{-6}}}



{{{m=(-6)/(2)}}} Subtract {{{-3}}} from {{{-1}}} to get {{{2}}}



{{{m=-3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,6\right)] and *[Tex \LARGE \left(-1,0\right)] is {{{m=-3}}}



Now let's use the point slope formula:



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



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



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



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



{{{y-6=-3x-9}}} Multiply



{{{y=-3x-9+6}}} Add 6 to both sides. 



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



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



So the equation that goes through the points *[Tex \LARGE \left(-3,6\right)] and *[Tex \LARGE \left(-1,0\right)] is {{{y=-3x-3}}}



 Notice how the graph of {{{y=-3x-3}}} goes through the points *[Tex \LARGE \left(-3,6\right)] and *[Tex \LARGE \left(-1,0\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-3x-3),
 circle(-3,6,0.08),
 circle(-3,6,0.10),
 circle(-3,6,0.12),
 circle(-1,0,0.08),
 circle(-1,0,0.10),
 circle(-1,0,0.12)
 )}}} Graph of {{{y=-3x-3}}} through the points *[Tex \LARGE \left(-3,6\right)] and *[Tex \LARGE \left(-1,0\right)]