Question 202034


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



{{{y-3=(4)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



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


{{{y-3=4x+8}}} Multiply {{{4}}} and {{{2}}} to get {{{8}}}


{{{y=4x+8+3}}} Add 3 to  both sides to isolate y


{{{y=4x+11}}} Combine like terms {{{8}}} and {{{3}}} to get {{{11}}} 

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Answer:



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


{{{y=4x+11}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=4}}} and the y-intercept is {{{b=11}}}


Notice if we graph the equation {{{y=4x+11}}} and plot the point ({{{-2}}},{{{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, -11, 7, -6, 12,
graph(500, 500, -11, 7, -6, 12,(4)x+11),
circle(-2,3,0.12),
circle(-2,3,0.12+0.03)
) }}} Graph of {{{y=4x+11}}} through the point ({{{-2}}},{{{3}}})

and 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 ({{{-2}}},{{{3}}}), this verifies our answer.