\(k=tan\ a\)
\(k=\frac{y_2-y_1}{x_2-x_1}\)
\(l_1//l_2\Longleftrightarrow\ k_1\ =k_2\)
由
\(tan\ a_2=tan(90^\circ+a_1)=-\frac{1}{tan\ a_1}\)
得
\(k_1k_2=-1\)
即
\(l_1\bot l_2\Longleftrightarrow\ k_1k_2=-1\)
\(k=\frac{y-y_0}{x-x_0}\)
即
\(y-y_0=k(x-x_0)\)
\(y=kx+b\)
原文:https://www.cnblogs.com/kingBook/p/10703795.html