Question 115225
#1




If you want to find the equation of line with a given a slope of {{{-3/4}}} which goes through the point ({{{8}}},{{{2}}}), 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-2=(-3/4)(x-8)}}} Plug in {{{m=-3/4}}}, {{{x[1]=8}}}, and {{{y[1]=2}}} (these values are given)



{{{y-2=(-3/4)x+(-3/4)(-8)}}} Distribute {{{-3/4}}}


{{{y-2=(-3/4)x+6}}} Multiply {{{-3/4}}} and {{{-8}}} to get {{{6}}}


{{{y=(-3/4)x+6+2}}} Add 2 to  both sides to isolate y


{{{y=(-3/4)x+8}}} Combine like terms {{{6}}} and {{{2}}} to get {{{8}}} 

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Answer:



So the equation of the line with a slope of {{{-3/4}}} which goes through the point ({{{8}}},{{{2}}}) is:


{{{y=(-3/4)x+8}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-3/4}}} and the y-intercept is {{{b=8}}}


Notice if we graph the equation {{{y=(-3/4)x+8}}} and plot the point ({{{8}}},{{{2}}}),  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, -1, 17, -7, 11,
graph(500, 500, -1, 17, -7, 11,(-3/4)x+8),
circle(8,2,0.12),
circle(8,2,0.12+0.03)
) }}} Graph of {{{y=(-3/4)x+8}}} through the point ({{{8}}},{{{2}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-3/4}}} and goes through the point ({{{8}}},{{{2}}}), this verifies our answer.


<hr>


#2




If you want to find the equation of line with a given a slope of {{{-1/2}}} which goes through the point ({{{4}}},{{{-5}}}), 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--5=(-1/2)(x-4)}}} Plug in {{{m=-1/2}}}, {{{x[1]=4}}}, and {{{y[1]=-5}}} (these values are given)



{{{y+5=(-1/2)(x-4)}}} Rewrite {{{y--5}}} as {{{y+5}}}



{{{y+5=(-1/2)x+(-1/2)(-4)}}} Distribute {{{-1/2}}}


{{{y+5=(-1/2)x+2}}} Multiply {{{-1/2}}} and {{{-4}}} to get {{{2}}}


{{{y=(-1/2)x+2-5}}} Subtract 5 from  both sides to isolate y


{{{y=(-1/2)x-3}}} Combine like terms {{{2}}} and {{{-5}}} to get {{{-3}}} 

------------------------------------------------------------------------------------------------------------

Answer:



So the equation of the line with a slope of {{{-1/2}}} which goes through the point ({{{4}}},{{{-5}}}) is:


{{{y=(-1/2)x-3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-1/2}}} and the y-intercept is {{{b=-3}}}


Notice if we graph the equation {{{y=(-1/2)x-3}}} and plot the point ({{{4}}},{{{-5}}}),  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, -5, 13, -14, 4,
graph(500, 500, -5, 13, -14, 4,(-1/2)x+-3),
circle(4,-5,0.12),
circle(4,-5,0.12+0.03)
) }}} Graph of {{{y=(-1/2)x-3}}} through the point ({{{4}}},{{{-5}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-1/2}}} and goes through the point ({{{4}}},{{{-5}}}), this verifies our answer.



<hr>


#3


If we know that the y-intercept is -3, then the line goes through (0,-3)





If you want to find the equation of line with a given a slope of {{{2}}} which goes through the point ({{{0}}},{{{-3}}}), 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--3=(2)(x-0)}}} Plug in {{{m=2}}}, {{{x[1]=0}}}, and {{{y[1]=-3}}} (these values are given)



{{{y+3=(2)(x-0)}}} Rewrite {{{y--3}}} as {{{y+3}}}



{{{y+3=2x+(2)(-0)}}} Distribute {{{2}}}


{{{y+3=2x+0}}} Multiply {{{2}}} and {{{-0}}} to get {{{0}}}


{{{y=2x+0-3}}} Subtract 3 from  both sides to isolate y


{{{y=2x-3}}} Combine like terms {{{0}}} and {{{-3}}} to get {{{-3}}} 

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Answer:



So the equation of the line with a slope of {{{2}}} which goes through the point ({{{0}}},{{{-3}}}) is:


{{{y=2x-3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2}}} and the y-intercept is {{{b=-3}}}


Notice if we graph the equation {{{y=2x-3}}} and plot the point ({{{0}}},{{{-3}}}),  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, -9, 9, -12, 6,
graph(500, 500, -9, 9, -12, 6,(2)x+-3),
circle(0,-3,0.12),
circle(0,-3,0.12+0.03)
) }}} Graph of {{{y=2x-3}}} through the point ({{{0}}},{{{-3}}})

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 ({{{0}}},{{{-3}}}), this verifies our answer.