Question 115201
There are a lot of problems here. So I'll do a few to help you get started

#1



If you want to find the equation of line with a given a slope of {{{2}}} which goes through the point ({{{-3}}},{{{1}}}), you can simply use the point-slope formula to find the equation:



---Point-Slope Formula---
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point


So lets use the Point-Slope Formula to find the equation of the line


{{{y-1=(2)(x--3)}}} Plug in {{{m=2}}}, {{{x[1]=-3}}}, and {{{y[1]=1}}} (these values are given)



{{{y-1=(2)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y-1=2x+(2)(3)}}} Distribute {{{2}}}


{{{y-1=2x+6}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}


{{{y=2x+6+1}}} Add 1 to  both sides to isolate y


{{{y=2x+7}}} Combine like terms {{{6}}} and {{{1}}} to get {{{7}}} 

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Answer:



So the equation of the line with a slope of {{{2}}} which goes through the point ({{{-3}}},{{{1}}}) is:


{{{y=2x+7}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2}}} and the y-intercept is {{{b=7}}}


Notice if we graph the equation {{{y=2x+7}}} and plot the point ({{{-3}}},{{{1}}}),  we get (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)


{{{drawing(500, 500, -12, 6, -8, 10,
graph(500, 500, -12, 6, -8, 10,(2)x+7),
circle(-3,1,0.12),
circle(-3,1,0.12+0.03)
) }}} Graph of {{{y=2x+7}}} through the point ({{{-3}}},{{{1}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{2}}} and goes through the point ({{{-3}}},{{{1}}}), this verifies our answer.


<hr>


#2




If you want to find the equation of line with a given a slope of {{{4}}} which goes through the point ({{{-6}}},{{{1}}}), you can simply use the point-slope formula to find the equation:



---Point-Slope Formula---
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point


So lets use the Point-Slope Formula to find the equation of the line


{{{y-1=(4)(x--6)}}} Plug in {{{m=4}}}, {{{x[1]=-6}}}, and {{{y[1]=1}}} (these values are given)



{{{y-1=(4)(x+6)}}} Rewrite {{{x--6}}} as {{{x+6}}}



{{{y-1=4x+(4)(6)}}} Distribute {{{4}}}


{{{y-1=4x+24}}} Multiply {{{4}}} and {{{6}}} to get {{{24}}}


{{{y=4x+24+1}}} Add 1 to  both sides to isolate y


{{{y=4x+25}}} Combine like terms {{{24}}} and {{{1}}} to get {{{25}}} 

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Answer:



So the equation of the line with a slope of {{{4}}} which goes through the point ({{{-6}}},{{{1}}}) is:


{{{y=4x+25}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=4}}} and the y-intercept is {{{b=25}}}


Notice if we graph the equation {{{y=4x+25}}} and plot the point ({{{-6}}},{{{1}}}),  we get (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)


{{{drawing(500, 500, -15, 3, -8, 10,
graph(500, 500, -15, 3, -8, 10,(4)x+25),
circle(-6,1,0.12),
circle(-6,1,0.12+0.03)
) }}} Graph of {{{y=4x+25}}} through the point ({{{-6}}},{{{1}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{4}}} and goes through the point ({{{-6}}},{{{1}}}), this verifies our answer.


<hr>


#3


If you want to find the equation of line with an undefined slope which goes through the point ({{{10}}},{{{4}}}), simply draw a vertical line through {{{x=10}}}:


{{{drawing(500, 500, -2, 19, -5, 13,
graph(500, 500, -2, 19, -5, 13,0),
line(10,-20,10,20)
) }}} Graph of {{{x=10}}}


So this simply means that the equation of the line is {{{x=10}}} (notice y is not in the equation. So this means this equation is <font size="4"><b>not</b></font> in slope-intercept form)