Skorzystamy z twierdzenia o współczynniku kierunkowym:

Jeśli dwa różne punkty A(x_A, y_A) i B(x_B, y_B) należą do wykresu funkcji liniowej y=ax+b, to

$a=\frac{y_B-y_A}{x_B-x_A}$  

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$\text{a)}A\left(-8,9\right),B\left(4,-6\right)$  

$x_A=-8,y_A=9$  

$x_B=4,y_B=-6$  

Mamy więc:

$a=\frac{y_B-y_A}{x_B-x_A}=\frac{-6-9}{4-\left(-8\right)}=\frac{-15}{12}=-\frac{5}{4}=-1\frac{1}{4}$  

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$\text{b)}A\left(26,-10\right),B\left(6,14\right)$  

$x_A=26,y_A=-10$  

$x_B=6,y_B=14$  

Mamy więc:

$a=\frac{y_B-y_A}{x_B-x_A}=\frac{14-\left(-10\right)}{6-26}=\frac{24}{-20}=-\frac{6}{5}=-1\frac{1}{5}$  

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$\text{c)}A\left(4,7\right),B\left(-6,-12\right)$  

$x_A=4,y_A=7$  

$x_B=-6,y_B=-12$  

Mamy więc:

$a=\frac{y_B-y_A}{x_B-x_A}=\frac{-12-7}{-6-4}=\frac{-19}{-10}=\frac{19}{10}=1,9$  

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$\text{d)}A\left(36,18\right),B\left(-4,-2\right)$  

$x_A=36,y_A=18$  

$x_B=-4,y_B=-2$  

Mamy więc:

$a=\frac{y_B-y_A}{x_B-x_A}=\frac{-2-18}{-4-36}=\frac{-20}{-40}=\frac{1}{2}$  

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$\text{e)}A\left(8,\sqrt{3}\right),B\left(9,\sqrt{3}\right)$  

$x_A=8,y_A=\sqrt{3}$  

$x_B=9,y_B=\sqrt{3}$  

Mamy więc:

$a=\frac{y_B-y_A}{x_B-x_A}=\frac{\sqrt{3}-\sqrt{3}}{9-8}=\frac{0}{1}=0$  

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$\text{f)}A\left(-5,8\frac{3}{4}\right),B\left(1,9\frac{1}{4}\right)$  

$x_A=-5,y_A=8\frac{3}{4}$  

$x_B=1,y_B=9\frac{1}{4}$  

Mamy więc:

$a=\frac{y_B-y_A}{x_B-x_A}=\frac{9\frac{1}{4}-8\frac{3}{4}}{1-\left(-5\right)}=\frac{\frac{2}{4}}{6}=\frac{1}{2}\cdot \frac{1}{6}=\frac{1}{12}$