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Boa noite Marcelo!
Solução!
Conhecendo os pontos,vamos primeiro determinar o coeficiente angular da reta.
![A(-3,2)\\\\\
B(0,3)\\\\\\
m= \dfrac{yB-yA}{xB-xA}\\\\\\
m= \dfrac{3-2}{0+3}\\\\\\
m= \dfrac{1}{3}](https://tex.z-dn.net/?f=A%28-3%2C2%29%5C%5C%5C%5C%5C%0AB%280%2C3%29%5C%5C%5C%5C%5C%5C%0Am%3D+%5Cdfrac%7ByB-yA%7D%7BxB-xA%7D%5C%5C%5C%5C%5C%5C%0A+m%3D+%5Cdfrac%7B3-2%7D%7B0%2B3%7D%5C%5C%5C%5C%5C%5C%0Am%3D+%5Cdfrac%7B1%7D%7B3%7D+ "A(-3,2)\\\\\
B(0,3)\\\\\\
m= \dfrac{yB-yA}{xB-xA}\\\\\\
m= \dfrac{3-2}{0+3}\\\\\\
m= \dfrac{1}{3}")
Equação da reta!
![y-yA=m(x-xA)\\\\\\\
y-2= \frac{1}{3}(x+3)\\\\\
y-2= \frac{x}{3}+1\\\\\
3y-6=x+3\\\\\
3y=x+3+6\\\\\\
3y=x+9\\\\\
\boxed{y= \frac{x}{3}+3}](https://tex.z-dn.net/?f=y-yA%3Dm%28x-xA%29%5C%5C%5C%5C%5C%5C%5C%0Ay-2%3D+%5Cfrac%7B1%7D%7B3%7D%28x%2B3%29%5C%5C%5C%5C%5C%0Ay-2%3D+%5Cfrac%7Bx%7D%7B3%7D%2B1%5C%5C%5C%5C%5C%0A3y-6%3Dx%2B3%5C%5C%5C%5C%5C%0A3y%3Dx%2B3%2B6%5C%5C%5C%5C%5C%5C%0A3y%3Dx%2B9%5C%5C%5C%5C%5C%0A%5Cboxed%7By%3D+%5Cfrac%7Bx%7D%7B3%7D%2B3%7D+++ "y-yA=m(x-xA)\\\\\\\
y-2= \frac{1}{3}(x+3)\\\\\
y-2= \frac{x}{3}+1\\\\\
3y-6=x+3\\\\\
3y=x+3+6\\\\\\
3y=x+9\\\\\
\boxed{y= \frac{x}{3}+3}")
Boa noite!
Bons estudos!
Solução!
Conhecendo os pontos,vamos primeiro determinar o coeficiente angular da reta.
Equação da reta!
Boa noite!
Bons estudos!
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