Question 117172
Since the equation has the x-intercept -3 and the y-intercept -5, this means the equation goes through the points (-3,0) and (0,-5)



First lets find the slope through the points ({{{-3}}},{{{0}}}) and ({{{0}}},{{{-5}}})


{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{-3}}},{{{0}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{0}}},{{{-5}}}))


{{{m=(-5-0)/(0--3)}}} Plug in {{{y[2]=-5}}},{{{y[1]=0}}},{{{x[2]=0}}},{{{x[1]=-3}}}  (these are the coordinates of given points)


{{{m= -5/3}}} Subtract the terms in the numerator {{{-5-0}}} to get {{{-5}}}.  Subtract the terms in the denominator {{{0--3}}} to get {{{3}}}

  

So the slope is

{{{m=-5/3}}}


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Now let's use the point-slope formula to find the equation of the line:




------Point-Slope Formula------
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(\textrm{x_{1},y_{1}}\right)] is one of the given points


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


{{{y-0=(-5/3)(x--3)}}} Plug in {{{m=-5/3}}}, {{{x[1]=-3}}}, and {{{y[1]=0}}} (these values are given)



{{{y-0=(-5/3)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y-0=(-5/3)x+(-5/3)(3)}}} Distribute {{{-5/3}}}


{{{y-0=(-5/3)x-5}}} Multiply {{{-5/3}}} and {{{3}}} to get {{{-15/3}}}. Now reduce {{{-15/3}}} to get {{{-5}}}


{{{y=(-5/3)x-5+0}}} Add {{{0}}} to  both sides to isolate y


{{{y=(-5/3)x-5}}} Combine like terms {{{-5}}} and {{{0}}} to get {{{-5}}} 

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Answer:



So the equation of the line which goes through the points ({{{-3}}},{{{0}}}) and ({{{0}}},{{{-5}}})  is:{{{y=(-5/3)x-5}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=-5/3}}} and the y-intercept is {{{b=-5}}}


Notice if we graph the equation {{{y=(-5/3)x-5}}} and plot the points ({{{-3}}},{{{0}}}) and ({{{0}}},{{{-5}}}),  we get this: (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, -10.5, 7.5, -11.5, 6.5,
graph(500, 500, -10.5, 7.5, -11.5, 6.5,(-5/3)x+-5),
circle(-3,0,0.12),
circle(-3,0,0.12+0.03),
circle(0,-5,0.12),
circle(0,-5,0.12+0.03)
) }}} Graph of {{{y=(-5/3)x-5}}} through the points ({{{-3}}},{{{0}}}) and ({{{0}}},{{{-5}}})


Notice how the two points lie on the line. This graphically verifies our answer.



Now let's convert the equation into standard form



*[invoke converting_linear_equations "slope-intercept_to_standard", 1, 2, 3, "-5/3", -5]



So the standard equation that x-intercept -3 and the y-intercept -5 is 

{{{5x+3y=-15}}}