The Slope-Intercept Form of the equation of a line is:
[tex]y=mx+b[/tex]Where "m" is the slope and "b" is the y-intercept.
So, given the points:
[tex]\begin{gathered} \mleft(-4,3\mright) \\ \mleft(3,1\mright) \end{gathered}[/tex]You can find the slope by using this formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]In this case, you can set up that.
[tex]\begin{gathered} y_2=3 \\ y_1=1 \\ \\ x_2=-4 \\ x_1=3 \end{gathered}[/tex]Then, substituting values into the formula and evaluating, you get:
[tex]m=\frac{3-1}{-4-3}=\frac{2}{-7}=-\frac{2}{7}[/tex]Now you can substitute the slope of the line and the coordinate of one of the points given in the exercise, into the equation
[tex]y=mx+b[/tex]Then, substituting the coordinates of the point:
[tex](3,1)[/tex]And then solving for "b", you get that this is:
[tex]\begin{gathered} 1=(-\frac{2}{7})(3)+b \\ \\ 1=-\frac{6}{7}+b \\ \\ 1+\frac{6}{7}=b \\ \\ b=\frac{13}{7} \end{gathered}[/tex]Finally, knowing "m" and "b", you can determine that the equation of this line (in terms of "x"), is:
[tex]\begin{gathered} y=mx+b \\ \\ y=-\frac{2}{7}x+\frac{13}{7} \end{gathered}[/tex]The answer is:
[tex]y=-\frac{2}{7}x+\frac{13}{7}[/tex]