Odpowiedź :
[tex]\left \{ {{4=1a+b} \atop {5= 3a+b}} \right. \\\left \{ {{-4=-1a-b} \atop {5= 3a+b}} \right. \\----------------\\2a = 1\\a = \frac{1}{2}[/tex]
[tex]Zadanie\\\\A=(1,4)\ \ \ B=(3,5)\\\\Wspolczynnik\ kierunkowy\ a\\\\a=\frac{y_B-y_A}{x_B-x_A}\\\\a=\frac{5-4}{3-1}\\\\a=\frac{1}{2}\\\\Odp.\ "1"[/tex]