Question 226454


First let's find the slope of the line through the points *[Tex \LARGE \left(-2,-9\right)] and *[Tex \LARGE \left(0,-6\right)]



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(-6--9)/(0--2)}}} Plug in {{{y[2]=-6}}}, {{{y[1]=-9}}}, {{{x[2]=0}}}, and {{{x[1]=-2}}}



{{{m=(3)/(0--2)}}} Subtract {{{-9}}} from {{{-6}}} to get {{{3}}}



{{{m=(3)/(2)}}} Subtract {{{-2}}} from {{{0}}} to get {{{2}}}



So the slope of the line that goes through the points *[Tex \LARGE \left(-2,-9\right)] and *[Tex \LARGE \left(0,-6\right)] is {{{m=3/2}}}



Now let's use {{{y=mx+b}}} and {{{m=3/2}}} to find 'b'



{{{y=mx+b}}} Start with the general slope-intercept equation. 



{{{y=(3/2)x+b}}} Plug in the given slope {{{m=3/2}}}.



{{{-9=3/2(-2)+b}}} Plug in the {{{x=-2}}} and {{{y=-9}}} from the given point *[Tex \LARGE \left(-2,-9\right)].



{{{-9=-3+b}}}  <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">Multiply</a> {{{3/2}}} and {{{-2}}} to get {{{-3}}} 



{{{-9+3=b}}} Add {{{3}}} to both sides.



{{{-6=b}}} Combine like terms.



So the y-intercept is {{{b=-6}}}.



So the equation (in slope-intercept form) that goes through the points *[Tex \LARGE \left(-2,-9\right)] and *[Tex \LARGE \left(0,-6\right)] is {{{y=(3/2)x-6}}}



If you need to convert it into standard form, then...



{{{y=(3/2)x-6}}} Start with the slope-intercept equation.



{{{2y=3x-12}}} Multiply EVERY term by the LCD 2 to clear the fraction.



{{{-3x+2y=-12}}} Subtract 3x from both sides.



{{{3x-2y=12}}} Multiply EVERY term by -1 to make the x coefficient positive.



So the equation in standard form that goes through (-2,-9) and (0,-6) is {{{3x-2y=12}}}



Here's a graph to visually confirm our answer:



{{{drawing(500,500,-12,8,-19,1,
grid(1),
graph(500,500,-12,8,-19,1,(3/2)x-6),
circle(-2,-9,0.05),circle(-2,-9,0.08),circle(-2,-9,0.10),
circle(-2,-9,0.12),circle(-2,-9,0.15)
)}}}