Question 142365


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



{{{y-1=3x+(3)(-2)}}} Distribute {{{3}}}


{{{y-1=3x-6}}} Multiply {{{3}}} and {{{-2}}} to get {{{-6}}}


{{{y=3x-6+1}}} Add 1 to  both sides to isolate y


{{{y=3x-5}}} Combine like terms {{{-6}}} and {{{1}}} to get {{{-5}}} 

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

Answer:



So the equation of the line with a slope of {{{3}}} which goes through the point ({{{2}}},{{{1}}}) is:


{{{y=3x-5}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=3}}} and the y-intercept is {{{b=-5}}}


Notice if we graph the equation {{{y=3x-5}}} and plot the point ({{{2}}},{{{1}}}),  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, -7, 11, -8, 10,
graph(500, 500, -7, 11, -8, 10,(3)x+-5),
circle(2,1,0.12),
circle(2,1,0.12+0.03)
) }}} Graph of {{{y=3x-5}}} through the point ({{{2}}},{{{1}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{3}}} and goes through the point ({{{2}}},{{{1}}}), this verifies our answer.