Question Number 19505 by Tinkutara last updated on 12/Aug/17
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{two}\:\mathrm{straight}\:\mathrm{lines}\:\mathrm{with} \\ $$$$\mathrm{complex}\:\mathrm{slopes}\:\mu_{\mathrm{1}} \:\mathrm{and}\:\mu_{\mathrm{2}} \:\mathrm{are}\:\mathrm{parallel} \\ $$$$\mathrm{and}\:\mathrm{perpendicular}\:\mathrm{according}\:\mathrm{as}\:\mu_{\mathrm{1}} \:=\:\mu_{\mathrm{2}} \\ $$$$\mathrm{and}\:\mu_{\mathrm{1}} \:+\:\mu_{\mathrm{2}} \:=\:\mathrm{0}.\:\mathrm{Hence}\:\mathrm{if}\:\mathrm{the}\:\mathrm{straight} \\ $$$$\mathrm{lines}\:\bar {\alpha}{z}\:+\:\alpha\bar {{z}}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{and}\:\bar {\beta}{z}\:+\:\beta\bar {{z}}\:+\:{k}\:=\:\mathrm{0} \\ $$$$\mathrm{are}\:\mathrm{parallel}\:\mathrm{and}\:\mathrm{perpendicular}\:\mathrm{according} \\ $$$$\mathrm{as}\:\bar {\alpha}\beta\:−\:\alpha\bar {\beta}\:=\:\mathrm{0}\:\mathrm{and}\:\bar {\alpha}\beta\:+\:\alpha\bar {\beta}\:=\:\mathrm{0}. \\ $$
Answered by ajfour last updated on 12/Aug/17
$$\mathrm{complex}\:\mathrm{slope}\:\mu_{\mathrm{1}} =−\frac{\alpha}{\bar {\alpha}} \\ $$$$\:\mu_{\mathrm{2}} =−\frac{\beta}{\bar {\beta}}\: \\ $$$$\mathrm{so}\:\:\mu_{\mathrm{1}} =\mu_{\mathrm{2}} \:\:\:\Rightarrow\:\:\:\alpha\bar {\beta}−\bar {\alpha}\beta\:=\mathrm{0} \\ $$$$\mathrm{and}\:\:\:\mu_{\mathrm{1}} +\mu_{\mathrm{2}} =\mathrm{0}\:\:\Rightarrow\:\:\:\alpha\bar {\beta}+\bar {\alpha}\beta\:=\mathrm{0}\:. \\ $$
Commented by Tinkutara last updated on 12/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 12/Aug/17
$$\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{matter}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{slopes}.. \\ $$