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Sagot :
So we are given two points and we have to use them to construct the equation of a line in slope-intercept form:
[tex]y=mx+b[/tex]Where m is the slope and b is the y-intercept. With two given points we can construct two equations for m and b. For example, if we know that the line passes through two generic points (A,B) and (C,D) we have the following equations that are the result of taking (x,y)=(A,B) and (x,y)=(C,D):
[tex]\begin{gathered} B=A\cdot m+b \\ D=C\cdot m+b \end{gathered}[/tex]The points we have are (-4,-1) and (2,-5). Then we have A=-4, B=-1, C=2 and D=-5 and the two equations for m and b are:
[tex]\begin{gathered} -1=-4m+b \\ -5=2m+b \end{gathered}[/tex]We can take the first equation and add 4m to both sides:
[tex]\begin{gathered} -1+4m=-4m+b+4m \\ b=4m-1 \end{gathered}[/tex]Then we substitute b with this expression in the second equation:
[tex]\begin{gathered} -5=2m+b \\ -5=2m+4m-1 \\ -5=6m-1 \end{gathered}[/tex]And we add 1 to both sides:
[tex]\begin{gathered} -5+1=6m-1+1 \\ -4=6m \end{gathered}[/tex]And we divide both sides by 6:
[tex]\begin{gathered} -\frac{4}{6}=\frac{6m}{6} \\ m=-\frac{4}{6}=-\frac{2}{3} \end{gathered}[/tex]Then we use this value in the expression for b:
[tex]\begin{gathered} b=4m-1=4\cdot(-\frac{2}{3})-1 \\ b=-\frac{8}{3}-1 \\ b=-\frac{11}{3} \end{gathered}[/tex]Thent the equation we are looking for and answer to this question is:
[tex]y=-\frac{2}{3}x-\frac{11}{3}[/tex]
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