Question 92330
I'm assuming it suppose to look like "m=4" not "m+4"





If you want to find the equation of line with a given a slope of {{{4}}} which goes through the point ({{{3}}},{{{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 ({{{x[1]}}},{{{y[1]}}}) is the given point


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


{{{y-1=(4)(x-3)}}} Plug in {{{m=4}}}, {{{x[1]=3}}}, and {{{y[1]=1}}} (these values are given)


{{{y-1=(4)x-(4)(3))}}} Distribute {{{4}}}



{{{y-1=(4)x-12}}} Multiply {{{-4}}} and {{{3}}} to get {{{-12}}}


{{{y=(4)x-12+1}}}Add {{{1}}} to both sides


{{{y=4x-11}}} Combine like terms

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

Answer:



So the equation of the line with a slope of {{{4}}} which goes through the point ({{{3}}},{{{1}}}) 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 ({{{3}}},{{{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, -6, 12, -8, 10,
graph(500, 500, -6, 12, -8, 10,(4)x+-11),
circle(3,1,0.12),
circle(3,1,0.12+0.03)
) }}} Graph of {{{y=4x-11}}} through the point ({{{3}}},{{{1}}})

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 ({{{3}}},{{{1}}}), this verifies our answer.