Question 151492


If you want to find the equation of line with a given a slope of {{{4/3}}} which goes through the point ({{{-4}}},{{{-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 *[Tex \Large \left(x_{1},y_{1}\right)] is the given point


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


{{{y--5=(4/3)(x--4)}}} Plug in {{{m=4/3}}}, {{{x[1]=-4}}}, and {{{y[1]=-5}}} (these values are given)



{{{y+5=(4/3)(x--4)}}} Rewrite {{{y--5}}} as {{{y+5}}}



{{{y+5=(4/3)(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}



{{{y+5=(4/3)x+(4/3)(4)}}} Distribute {{{4/3}}}


{{{y+5=(4/3)x+16/3}}} Multiply {{{4/3}}} and {{{4}}} to get {{{16/3}}}


{{{y=(4/3)x+16/3-5}}} Subtract 5 from  both sides to isolate y


{{{y=(4/3)x+1/3}}} Combine like terms {{{16/3}}} and {{{-5}}} to get {{{1/3}}} (note: if you need help with combining fractions, check out this <a href=http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver>solver</a>)


{{{3y=4x+1}}} Multiply <b>every</b> term by the LCD 3 to clear the fractions.



{{{-4x+3y=1}}} Subtract {{{4x}}} from both sides.



{{{4x-3y=-1}}} Multiply every term by -1 (this will make the "x" coefficient positive).




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Answer:

So the equation in standard form that has a slope {{{m=4/3}}} and goes through the point (-4,-5) is


{{{4x-3y=-1}}}