find the slope of the line through each pair of points

The formula of a slope:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
95)
[tex](-13,\ -10)\to x_1=-13,\ y_1=-10\\(16,\ 16)\to x_2=16,\ y_2=16[/tex]
substitute:
[tex]m=\dfrac{16-(-10)}{16-(-13)}=\dfrac{16+10}{16+13}=\dfrac{26}{29}[/tex]
96)
[tex](6,\ -16)\to x_1=6,\ y_1=-16\\(-15,\ 9)\to x_2=-15,\ y_2=9[/tex]
substitute:
[tex]m=\dfrac{9-(-16)}{-15-6}=\dfrac{25}{-21}=-\dfrac{25}{21}[/tex]
97)
[tex](12,\ -11)\to x_1=12,\ y_1=-11\\(14,\ -12)\to x_2=14,\ y_2=-12[/tex]
substitute:
[tex]m=\dfrac{-12-(-11)}{14-12}=\dfrac{-12+11}{2}=-\dfrac{1}{2}[/tex]
OK. I think you already know how to solve it. I will give short solutions now.
98)
[tex]m=\dfrac{15-(-3)}{8-(-14)}=\dfrac{18}{22}=\dfrac{9}{11}[/tex]
99)
[tex]m=\dfrac{3-(-5)}{-7-3}=\dfrac{8}{-10}=-\dfrac{4}{5}[/tex]
100)
[tex]m=\dfrac{-8-8}{-11-0}=\dfrac{-16}{-11}=\dfrac{16}{11}[/tex]
101)
[tex]m=\dfrac{13-(-5)}{-13-(-13)}=\dfrac{18}{0}!!!-\boxed{NOT\ EXIST}[/tex]
102)
[tex]m=\dfrac{18-1}{-10-2}=\dfrac{17}{-12}=-\dfrac{17}{12}[/tex]
103)
[tex]m=\dfrac{19-14}{7-(-17)}=\dfrac{5}{24}[/tex]
104)
[tex]m=\dfrac{12-0}{-6-4}=\dfrac{12}{-10}=-\dfrac{6}{5}[/tex]