Question 243549


If you want to find the equation of line with a given a slope of {{{4}}} 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=(4)(x-0)}}} Plug in {{{m=4}}}, {{{x[1]=0}}}, and {{{y[1]=3}}} (these values are given)



{{{y-3=4x+(4)(0)}}} Distribute {{{4}}}



{{{y-3=4x+0}}} Multiply {{{4}}} and {{{0}}} to get {{{0}}}



{{{y=4x+0+3}}} Add 3 to  both sides to isolate y



{{{y=4x+3}}} Combine like terms.
 

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Answer:



So the equation of the line with a slope of {{{4}}} which goes through the point ({{{0}}},{{{3}}}) is:


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


Notice if we graph the equation {{{y=4x+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, -6, 12,
graph(500, 500, -9, 9, -6, 12,(4)x+3),
circle(0,3,0.12),
circle(0,3,0.12+0.03)
) }}} 


Graph of {{{y=4x+3}}} through the point (0,3)



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