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Answer:
[tex]y=-\frac{4}{5}x-\frac{8}{5}[/tex]First, let us find the slope of the line using the following equation:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Using the points (3, -4) and (8, -8)
[tex]m=\frac{y_2-y_1}{x_2-x_1}\Rightarrow m=\frac{-8-(-4)}{8-3}[/tex][tex]m=\frac{-8+4}{8-3}=\frac{-4}{5}\Rightarrow m=-\frac{4}{5}[/tex]Now that we found the slope of the line, we are going to use the following equation to solve for the equation of the line:
[tex]y-y_1=m(x-x_1)[/tex]Using the point (3, -4)
[tex]y-y_1=m(x-x_1)\Rightarrow y-(-4)=-\frac{4}{5}(x-3)[/tex][tex]y+4=-\frac{4}{5}x+\frac{12}{5}\Rightarrow y=-\frac{4}{5}x+\frac{12}{5}-4[/tex][tex]y=-\frac{4}{5}x-\frac{8}{5}[/tex]Therefore, the equation of the line that passes through the points (3, -4) and (8, -8) is:
[tex]y=-\frac{4}{5}x-\frac{8}{5}[/tex]
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