Question 90377


If you want to find the equation of line with a given a slope of {{{4}}} which goes through the point ({{{0}}},{{{-5}}}), 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--5=(4)(x-0)}}} Plug in {{{m=4}}}, {{{x[1]=0}}}, and {{{y[1]=-5}}} (these values are given)


{{{y--5=(4)x-(4)(0))}}} Distribute {{{4}}}


{{{y--5=(4)x+(-4)(0))}}} Multiply the negatives


{{{y+5=(4)x+0}}} Multiply {{{-4}}} and {{{0}}} to get {{{0}}}


{{{y=(4)x+0+-5}}}Subtract {{{-5}}} from both sides



{{{y=4x-5}}} 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}}},{{{-5}}}) is:


{{{y=4x-5}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=4}}} and the y-intercept is {{{b=-5}}}


Notice if we graph the equation {{{y=4x-5}}} and plot the point ({{{0}}},{{{-5}}}),  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, -14, 4,
graph(500, 500, -9, 9, -14, 4,(4)x+-5),
circle(0,-5,0.12),
circle(0,-5,0.12+0.03)
) }}} Graph of {{{y=4x-5}}} through the point ({{{0}}},{{{-5}}})

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 ({{{0}}},{{{-5}}}), this verifies our answer.