równanie prostej, uzasadnić wzór

[mateusz.ex]

Jeśli prosta \(\displaystyle{ y=ax+b}\) przechodzi przez dwa punkty \(\displaystyle{ A=( x_{1},y_{1})}\) oraz \(\displaystyle{ B=( x_{2},y _{2})}\) i \(\displaystyle{ x_{1}\neq x_{2}}\), to \(\displaystyle{ a= \frac{y _{2}-y _{1}}{x _{2}-x _{1}}}\)

[Nakahed90]

\(\displaystyle{ \begin{cases} y_{1}=ax_{1}+b\\y_{2}=ax_{2}+b\end{cases}}\)
Od drugiego odejmuje pierwsze.
\(\displaystyle{ y_{2}-y_{1}=ax_{2}-ax_{1}|:(x_{2}-x_{1}) \wedge x_{2}\neq x_{1}}\)
\(\displaystyle{ a=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}}\)

[marcinn12]

\(\displaystyle{ y _{1} =ax _{1} +b}\)
\(\displaystyle{ y _{2} =ax _{2} +b}\)
\(\displaystyle{ b=y _{1}-ax _{1}}\)
\(\displaystyle{ b=y _{2}-ax _{2}}\)
\(\displaystyle{ y _{1}-ax _{1}=y _{2}-ax _{2}}\)
\(\displaystyle{ ax_{2}-ax_{1}=y_{2}-y_{1}}\)
\(\displaystyle{ a(x_{2}-x_{1})=y_{2}-y_{1}}\)
\(\displaystyle{ a= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}}\)

Ale chyba się spoźniłem ;P