Question 750885


First let's find the slope of the line through the points *[Tex \LARGE \left(5,-2\right)] and *[Tex \LARGE \left(9,-2\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(5,-2\right)]. So this means that {{{x[1]=5}}} and {{{y[1]=-2}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(9,-2\right)].  So this means that {{{x[2]=9}}} and {{{y[2]=-2}}}.



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



{{{m=(-2--2)/(9-5)}}} Plug in {{{y[2]=-2}}}, {{{y[1]=-2}}}, {{{x[2]=9}}}, and {{{x[1]=5}}}



{{{m=(0)/(9-5)}}} Subtract {{{-2}}} from {{{-2}}} to get {{{0}}}



{{{m=(0)/(4)}}} Subtract {{{5}}} from {{{9}}} to get {{{4}}}



{{{m=0}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(5,-2\right)] and *[Tex \LARGE \left(9,-2\right)] is {{{m=0}}}



Now let's use the point slope formula:



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



{{{y--2=0(x-5)}}} Plug in {{{m=0}}}, {{{x[1]=5}}}, and {{{y[1]=-2}}}



{{{y+2=0(x-5)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=0x+0(-5)}}} Distribute



{{{y+2=0x+0}}} Multiply



{{{y=0x+0-2}}} Subtract 2 from both sides. 



{{{y=0x-2}}} Combine like terms. 



{{{y=-2}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(5,-2\right)] and *[Tex \LARGE \left(9,-2\right)] is {{{y=-2}}}



 Notice how the graph of {{{y=-2}}} goes through the points *[Tex \LARGE \left(5,-2\right)] and *[Tex \LARGE \left(9,-2\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-2),
 circle(5,-2,0.08),
 circle(5,-2,0.10),
 circle(5,-2,0.12),
 circle(9,-2,0.08),
 circle(9,-2,0.10),
 circle(9,-2,0.12)
 )}}} Graph of {{{y=-2}}} through the points *[Tex \LARGE \left(5,-2\right)] and *[Tex \LARGE \left(9,-2\right)]