Question 87742
#1


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


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


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


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


{{{y=(2)x+-2+-2}}}Subtract {{{-2}}} from both sides



{{{y=(2)x+-4}}} Combine like terms

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Answer:



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


{{{y=(2)x+-4}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=2}}} (so this shows our equation has a slope of {{{2}}}) and the y-intercept is {{{b=-4}}}


Notice if we graph the equation {{{y=(2)x+-4}}} (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>) and plot the point ({{{1}}},{{{-2}}}),  we get


{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(2)x+-4),
circle(1,-2,0.05),
circle(1,-2,0.08)
) }}} Graph of {{{y=(2)x+-4}}} through the point ({{{1}}},{{{-2}}})

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

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#2

*[invoke calculating_slope -1, 4, 2, -1]