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Sagot :
First, let's find the slope of the line that passes through the points (-4,3) and (6,2):
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ \Rightarrow m=\frac{2-3}{6-(-4)}=\frac{-1}{6+4}=-\frac{1}{10} \end{gathered}[/tex]Now we can use the first point to get the equation of the line:
[tex]\begin{gathered} (x_1,y_1)=(-4,3) \\ y-y_1=m(x-x_1) \\ \Rightarrow y-3=-\frac{1}{10}(x-(-4))=-\frac{1}{10}(x+4)=-\frac{1}{10}x-\frac{4}{10}=-\frac{1}{10}x-\frac{2}{5} \\ \Rightarrow y=-\frac{1}{10}x-\frac{2}{5}+3=-\frac{1}{10}x-\frac{2}{5}+\frac{15}{5}=-\frac{1}{10}x+\frac{13}{5} \\ y=-\frac{1}{10}x+\frac{13}{5} \end{gathered}[/tex]therefore, the equation of the line is y=-1/10x+13/5
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