Question 98452
1.


"Find an equation of the line with the given slope and through the given point.
1. slope=2;(3,11)"



If you want to find the equation of line with a given a slope of {{{2}}} which goes through the point ({{{3}}},{{{11}}}), 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-11=(2)(x-3)}}} Plug in {{{m=2}}}, {{{x[1]=3}}}, and {{{y[1]=11}}} (these values are given)



{{{y-11=2x+(2)(-3)}}} Distribute {{{2}}}


{{{y-11=2x-6}}} Multiply {{{2}}} and {{{-3}}} to get {{{-6}}}


{{{y=2x-6+11}}} Add 11 to  both sides to isolate y


{{{y=2x+5}}} Combine like terms {{{-6}}} and {{{11}}} to get {{{5}}} 

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Answer:



So the equation of the line with a slope of {{{2}}} which goes through the point ({{{3}}},{{{11}}}) is:


{{{y=2x+5}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2}}} and the y-intercept is {{{b=5}}}


Notice if we graph the equation {{{y=2x+5}}} and plot the point ({{{3}}},{{{11}}}),  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, -6, 12, 2, 20,
graph(500, 500, -6, 12, 2, 20,(2)x+5),
circle(3,11,0.12),
circle(3,11,0.12+0.03)
) }}} Graph of {{{y=2x+5}}} through the point ({{{3}}},{{{11}}})

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}}},{{{11}}}), this verifies our answer.



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2.


"Find an equation of the line through the given points.
(1,-5);(4,7)"



First lets find the slope through the points ({{{1}}},{{{-5}}}) and ({{{4}}},{{{7}}})


{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{1}}},{{{-5}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{4}}},{{{7}}}))


{{{m=(7--5)/(4-1)}}} Plug in {{{y[2]=7}}},{{{y[1]=-5}}},{{{x[2]=4}}},{{{x[1]=1}}}  (these are the coordinates of given points)


{{{m= 12/3}}} Subtract the terms in the numerator {{{7--5}}} to get {{{12}}}.  Subtract the terms in the denominator {{{4-1}}} to get {{{3}}}

  


{{{m=4}}} Reduce

  

So the slope is

{{{m=4}}}


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



Now let's use the point-slope formula to find the equation of the line:




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


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


{{{y--5=(4)(x-1)}}} Plug in {{{m=4}}}, {{{x[1]=1}}}, and {{{y[1]=-5}}} (these values are given)



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



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


{{{y+5=4x-4}}} Multiply {{{4}}} and {{{-1}}} to get {{{-4}}}


{{{y=4x-4-5}}} Subtract {{{5}}} from  both sides to isolate y


{{{y=4x-9}}} Combine like terms {{{-4}}} and {{{-5}}} to get {{{-9}}} 

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

Answer:



So the equation of the line which goes through the points ({{{1}}},{{{-5}}}) and ({{{4}}},{{{7}}})  is:{{{y=4x-9}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=4}}} and the y-intercept is {{{b=-9}}}


Notice if we graph the equation {{{y=4x-9}}} and plot the points ({{{1}}},{{{-5}}}) and ({{{4}}},{{{7}}}),  we get this: (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, -6.5, 11.5, -8, 10,
graph(500, 500, -6.5, 11.5, -8, 10,(4)x+-9),
circle(1,-5,0.12),
circle(1,-5,0.12+0.03),
circle(4,7,0.12),
circle(4,7,0.12+0.03)
) }}} Graph of {{{y=4x-9}}} through the points ({{{1}}},{{{-5}}}) and ({{{4}}},{{{7}}})


Notice how the two points lie on the line. This graphically verifies our answer.