Question 113991
#1




If you want to find the equation of line with a given a slope of {{{2}}} which goes through the point ({{{-2}}},{{{7}}}), 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-7=(2)(x--2)}}} Plug in {{{m=2}}}, {{{x[1]=-2}}}, and {{{y[1]=7}}} (these values are given)



{{{y-7=(2)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



{{{y-7=2x+(2)(2)}}} Distribute {{{2}}}


{{{y-7=2x+4}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}


{{{y=2x+4+7}}} Add 7 to  both sides to isolate y


{{{y=2x+11}}} Combine like terms {{{4}}} and {{{7}}} to get {{{11}}} 

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Answer:



So the equation of the line with a slope of {{{2}}} which goes through the point ({{{-2}}},{{{7}}}) is:


{{{y=2x+11}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2}}} and the y-intercept is {{{b=11}}}


Notice if we graph the equation {{{y=2x+11}}} and plot the point ({{{-2}}},{{{7}}}),  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, -11, 7, -2, 16,
graph(500, 500, -11, 7, -2, 16,(2)x+11),
circle(-2,7,0.12),
circle(-2,7,0.12+0.03)
) }}} Graph of {{{y=2x+11}}} through the point ({{{-2}}},{{{7}}})

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 ({{{-2}}},{{{7}}}), this verifies our answer.


<hr>


#2


Note: I'm going to write the slope -3.5 as {{{-7/2}}}





If you want to find the equation of line with a given a slope of {{{-7/2}}} which goes through the point ({{{-5}}},{{{-9}}}), 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--9=(-7/2)(x--5)}}} Plug in {{{m=-7/2}}}, {{{x[1]=-5}}}, and {{{y[1]=-9}}} (these values are given)



{{{y+9=(-7/2)(x--5)}}} Rewrite {{{y--9}}} as {{{y+9}}}



{{{y+9=(-7/2)(x+5)}}} Rewrite {{{x--5}}} as {{{x+5}}}



{{{y+9=(-7/2)x+(-7/2)(5)}}} Distribute {{{-7/2}}}


{{{y+9=(-7/2)x-35/2}}} Multiply {{{-7/2}}} and {{{5}}} to get {{{-35/2}}}


{{{y=(-7/2)x-35/2-9}}} Subtract 9 from  both sides to isolate y


{{{y=(-7/2)x-53/2}}} Combine like terms {{{-35/2}}} and {{{-9}}} to get {{{-53/2}}} (note: if you need help with combining fractions, check out this <a href=http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver>solver</a>)



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Answer:



So the equation of the line with a slope of {{{-7/2}}} which goes through the point ({{{-5}}},{{{-9}}}) is:


{{{y=(-7/2)x-53/2}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-7/2}}} and the y-intercept is {{{b=-53/2}}}


Notice if we graph the equation {{{y=(-7/2)x-53/2}}} and plot the point ({{{-5}}},{{{-9}}}),  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, -14, 4, -18, 0,
graph(500, 500, -14, 4, -18, 0,(-7/2)x+-53/2),
circle(-5,-9,0.12),
circle(-5,-9,0.12+0.03)
) }}} Graph of {{{y=(-7/2)x-53/2}}} through the point ({{{-5}}},{{{-9}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-7/2}}} and goes through the point ({{{-5}}},{{{-9}}}), this verifies our answer.


<hr>


#3




If you want to find the equation of line with a given a slope of {{{7}}} which goes through the point ({{{4}}},{{{0}}}), 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-0=(7)(x-4)}}} Plug in {{{m=7}}}, {{{x[1]=4}}}, and {{{y[1]=0}}} (these values are given)



{{{y-0=7x+(7)(-4)}}} Distribute {{{7}}}


{{{y-0=7x-28}}} Multiply {{{7}}} and {{{-4}}} to get {{{-28}}}


{{{y=7x-28+0}}} Add 0 to  both sides to isolate y


{{{y=7x-28}}} Combine like terms {{{-28}}} and {{{0}}} to get {{{-28}}} 

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Answer:



So the equation of the line with a slope of {{{7}}} which goes through the point ({{{4}}},{{{0}}}) is:


{{{y=7x-28}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=7}}} and the y-intercept is {{{b=-28}}}


Notice if we graph the equation {{{y=7x-28}}} and plot the point ({{{4}}},{{{0}}}),  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, -9, 9,
graph(500, 500, -5, 13, -9, 9,(7)x+-28),
circle(4,0,0.12),
circle(4,0,0.12+0.03)
) }}} Graph of {{{y=7x-28}}} through the point ({{{4}}},{{{0}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{7}}} and goes through the point ({{{4}}},{{{0}}}), this verifies our answer.