Question 551090
I'll do the first one to get you started.


# 1




First let's find the slope of the line through the points *[Tex \LARGE \left(0,-4\right)] and *[Tex \LARGE \left(-2,2\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(0,-4\right)]. So this means that {{{x[1]=0}}} and {{{y[1]=-4}}}.

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



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



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



{{{m=(6)/(-2-0)}}} Subtract {{{-4}}} from {{{2}}} to get {{{6}}}



{{{m=(6)/(-2)}}} Subtract {{{0}}} from {{{-2}}} to get {{{-2}}}



{{{m=-3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(0,-4\right)] and *[Tex \LARGE \left(-2,2\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--4=-3(x-0)}}} Plug in {{{m=-3}}}, {{{x[1]=0}}}, and {{{y[1]=-4}}}



{{{y+4=-3(x-0)}}} Rewrite {{{y--4}}} as {{{y+4}}}



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



{{{y+4=-3x+0}}} Multiply



{{{y=-3x+0-4}}} Subtract 4 from both sides. 



{{{y=-3x-4}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(0,-4\right)] and *[Tex \LARGE \left(-2,2\right)] is {{{y=-3x-4}}}



 Notice how the graph of {{{y=-3x-4}}} goes through the points *[Tex \LARGE \left(0,-4\right)] and *[Tex \LARGE \left(-2,2\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-3x-4),
 circle(0,-4,0.08),
 circle(0,-4,0.10),
 circle(0,-4,0.12),
 circle(-2,2,0.08),
 circle(-2,2,0.10),
 circle(-2,2,0.12)
 )}}} Graph of {{{y=-3x-4}}} through the points *[Tex \LARGE \left(0,-4\right)] and *[Tex \LARGE \left(-2,2\right)]