for 9 and 9A, find the slope of the line​

For this case we have that by definition, the slope of a line is given by:[tex]m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}}[/tex]Where:[tex](x_ {1}, y_ {1})[/tex] and [tex](x_ {2}, y_ {2})[/tex] are two points through which the line passes.Question 1:According to the image, we have the following points:[tex](x_ {1}, y_ {1}) :( 2, -6)\\(x_ {2}, y_ {2}): (- 6,5)[/tex]Substituting we have:[tex]m = \frac {5 - (- 6)} {- 6-2} = \frac {5 + 6} {- 8} = \frac {11} {- 8} = - \frac {11} {8}[/tex]Thus, the slope is: [tex]- \frac {11} {8}[/tex]Question 2:According to the image, the line goes through the following points:[tex](x_ {1}, y_ {1}): (- 1,2)\\(x_ {2}, y_ {2}) :( 2, -2)[/tex]Substituting we have:[tex]m = \frac {-2-2} {2 - (- 1)} = \frac {-4} {2 + 1} = \frac {-4} {3} = - \frac {4} {3}[/tex]Thus, the slope is: [tex]- \frac {4} {3}[/tex]Answer:Slope 1: [tex]- \frac {11} {8}[/tex]Slope 2:[tex]- \frac {4} {3}[/tex]