Question 89012


If you want to find the equation of line with a given a slope of {{{-1/2}}} 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 ({{{x[1]}}},{{{y[1]}}}) is the given point


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


{{{y-3=(-1/2)(x-0)}}} Plug in {{{m=-1/2}}}, {{{x[1]=0}}}, and {{{y[1]=3}}} (these values are given)


{{{y-3=(-1/2)x-(-1/2)(0))}}} Distribute {{{-1/2}}}


{{{y-3=(-1/2)x+(1/2)(0))}}} Multiply the negatives


{{{y-3=(-1/2)x+0}}} Multiply {{{1/2}}} and {{{0}}} to get {{{0}}}


{{{y=(-1/2)x+0+3}}}Add {{{3}}} to both sides


{{{y=(-1/2)x+3}}} Combine like terms

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Answer:



So the equation of the line with a slope of {{{-1/2}}} which goes through the point ({{{0}}},{{{3}}}) is:


{{{y=(-1/2)x+3}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-1/2}}} and the y-intercept is {{{b=3}}}


Notice if we graph the equation {{{y=(-1/2)x+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,(-1/2)x+3),
circle(0,3,0.12),
circle(0,3,0.12+0.03)
) }}} Graph of {{{y=(-1/2)x+3}}} through the point ({{{0}}},{{{3}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-1/2}}} and goes through the point ({{{0}}},{{{3}}}), this verifies our answer.