Odpowiedź:
[tex]R\'ownanie\ \ prostej\ \ w\ \ postaci\ \ kierunkowej\ \ to\ \ y=ax+b\\\\Podstawiamy\ \ wsp\'olrzedne\ \ punkt\'ow\ \ A\ \ i\ \ B\ \ do\ \ wzoru\ \ i\ \ tworzymy\ \ uklad\ \ r\'owna\'n\\\\\begin{cases}3=a\cdot(-1)+b\\1=a\cdot5+b\end{cases}\\\\\\\begin{cases}3=-a+b\ \ /\cdot(-1)\\1=5a+b\end{cases}\\\\\\+\begin{cases}-3=a-b\\1=5a+b\end{cases}\\--------\\-2=6a\\\\6a=-2\ \ /:6\\\\a=-\frac{2}{6}\\\\a=-\frac{1}{3}\\\\Podstawiamy\ \ warto\'s\'c\ \ a=-\frac{1}{3}\ \ do\ \ drugiego\ \ r\'ownania[/tex]
[tex]1=5\cdot(-\frac{1}{3})+b\\\\1=-\frac{5}{3}+b\\\\1+\frac{5}{3}=b\\\\\frac{8}{3}=b\\\\b=\frac{8}{3}\\\\R\'ownanie\ \ prostej\ \ ma\ \ posta\'c\ \ y=-\frac{1}{3}x+\frac{8}{3}[/tex]
