С(x_(3);y_(3))
D(x_(4);y_(4))
vector{CA}=(x_(3)-x_(1);y_(3)-y_(1))
vector{DB}=(x_(4)-x_(2);y_(4)-y_(2))
Пусть СА || оси Ох
vector{CA} коллинеарен вектору vector{i}=(1;0)
координаты коллинеарных векторов пропорциональны
x_(3)-x_(1)=k ⇒ [b]x_(3)[/b] = x_(1)+k
y_(3)-y_(1)=0 ⇒ [b]y_(3)[/b] = y_(1)
Аналогично
x_(4)-x_(2)=0⇒ [b]x_(4)[/b] = x_(2)
y_(4)-y_(2)=k ⇒ [b]y_(4)[/b] = y_(2)+k