Question 169806


First let's find the slope of the line through the points *[Tex \LARGE \left(-1,-2\right)] and *[Tex \LARGE \left(-3,0\right)]



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



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



{{{m=(2)/(-3--1)}}} Subtract {{{-2}}} from {{{0}}} to get {{{2}}}



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



{{{m=-1}}} Reduce



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



{{{y--2=-1(x+1)}}} Rewrite {{{x--1}}} as {{{x+1}}}



{{{y+2=-1(x+1)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=-1x+-1(1)}}} Distribute



{{{y+2=-1x-1}}} Multiply



{{{y=-1x-1-2}}} Subtract 2 from both sides. 



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



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



So the equation that goes through the points *[Tex \LARGE \left(-1,-2\right)] and *[Tex \LARGE \left(-3,0\right)] is {{{y=-x-3}}}



 Notice how the graph of {{{y=-x-3}}} goes through the points *[Tex \LARGE \left(-1,-2\right)] and *[Tex \LARGE \left(-3,0\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-x-3),
 circle(-1,-2,0.08),
 circle(-1,-2,0.10),
 circle(-1,-2,0.12),
 circle(-3,0,0.08),
 circle(-3,0,0.10),
 circle(-3,0,0.12)
 )}}} Graph of {{{y=-x-3}}} through the points *[Tex \LARGE \left(-1,-2\right)] and *[Tex \LARGE \left(-3,0\right)]