Question 118103
#1


Anything parallel to {{{y=2x}}} will have a slope of 2. So just add any number (I'm going to add 2) to {{{y=2x}}} to get {{{y=2x+2}}}


Notice if we graph {{{y=2x}}} and {{{y=2x+2}}} we can see that the two lines are parallel. So this visually verifies our answer.


{{{ graph( 500, 500, -10, 10, -10, 10, 2x,2x+2) }}} Graph of {{{y=2x}}} (red) and {{{y=2x+2}}} (green)



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#2


Remember the origin is the point (0,0)



First lets find the slope through the points ({{{0}}},{{{0}}}) and ({{{1}}},{{{2}}})


{{{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 ({{{0}}},{{{0}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{1}}},{{{2}}}))


{{{m=(2-0)/(1-0)}}} Plug in {{{y[2]=2}}},{{{y[1]=0}}},{{{x[2]=1}}},{{{x[1]=0}}}  (these are the coordinates of given points)


{{{m= 2/1}}} Subtract the terms in the numerator {{{2-0}}} to get {{{2}}}.  Subtract the terms in the denominator {{{1-0}}} to get {{{1}}}

  


{{{m=2}}} Reduce

  

So the slope is

{{{m=2}}}


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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-0=(2)(x-0)}}} Plug in {{{m=2}}}, {{{x[1]=0}}}, and {{{y[1]=0}}} (these values are given)



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


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


{{{y=2x+0+0}}} Add {{{0}}} to  both sides to isolate y


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

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Answer:



So the equation of the line which goes through the points ({{{0}}},{{{0}}}) and ({{{1}}},{{{2}}})  is:{{{y=2x+0}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=2}}} and the y-intercept is {{{b=0}}}


Notice if we graph the equation {{{y=2x+0}}} and plot the points ({{{0}}},{{{0}}}) and ({{{1}}},{{{2}}}),  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, -8.5, 9.5, -8, 10,
graph(500, 500, -8.5, 9.5, -8, 10,(2)x+0),
circle(0,0,0.12),
circle(0,0,0.12+0.03),
circle(1,2,0.12),
circle(1,2,0.12+0.03)
) }}} Graph of {{{y=2x+0}}} through the points ({{{0}}},{{{0}}}) and ({{{1}}},{{{2}}})


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