Question 367354


First let's find the slope of the line through the points *[Tex \LARGE \left(-2,0\right)] and *[Tex \LARGE \left(0,3\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-2,0\right)]. So this means that {{{x[1]=-2}}} and {{{y[1]=0}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(0,3\right)].  So this means that {{{x[2]=0}}} and {{{y[2]=3}}}.



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



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



{{{m=(3)/(0--2)}}} Subtract {{{0}}} from {{{3}}} 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(-2,0\right)] and *[Tex \LARGE \left(0,3\right)] is {{{m=3/2}}}



Now let's use the point slope formula:



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



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



{{{y-0=(3/2)(x+2)}}} Rewrite {{{x--2}}} as {{{x+2}}}



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



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



{{{y=(3/2)x+3}}} Simplify



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



 Notice how the graph of {{{y=(3/2)x+3}}} goes through the points *[Tex \LARGE \left(-2,0\right)] and *[Tex \LARGE \left(0,3\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(3/2)x+3),
 circle(-2,0,0.08),
 circle(-2,0,0.10),
 circle(-2,0,0.12),
 circle(0,3,0.08),
 circle(0,3,0.10),
 circle(0,3,0.12)
 )}}} Graph of {{{y=(3/2)x+3}}} through the points *[Tex \LARGE \left(-2,0\right)] and *[Tex \LARGE \left(0,3\right)]

 

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