Menu Close

Prove-that-two-straight-lines-with-complex-slopes-1-and-2-are-parallel-and-perpendicular-according-as-1-2-and-1-2-0-Hence-if-the-straight-lines-z-z-c-0-and-z




Question Number 19505 by Tinkutara last updated on 12/Aug/17
Prove that two straight lines with  complex slopes μ_1  and μ_2  are parallel  and perpendicular according as μ_1  = μ_2   and μ_1  + μ_2  = 0. Hence if the straight  lines α^� z + αz^�  + c = 0 and β^� z + βz^�  + k = 0  are parallel and perpendicular according  as α^� β − αβ^�  = 0 and α^� β + αβ^�  = 0.
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{two}\:\mathrm{straight}\:\mathrm{lines}\:\mathrm{with} \\ $$$$\mathrm{complex}\:\mathrm{slopes}\:\mu_{\mathrm{1}} \:\mathrm{and}\:\mu_{\mathrm{2}} \:\mathrm{are}\:\mathrm{parallel} \\ $$$$\mathrm{and}\:\mathrm{perpendicular}\:\mathrm{according}\:\mathrm{as}\:\mu_{\mathrm{1}} \:=\:\mu_{\mathrm{2}} \\ $$$$\mathrm{and}\:\mu_{\mathrm{1}} \:+\:\mu_{\mathrm{2}} \:=\:\mathrm{0}.\:\mathrm{Hence}\:\mathrm{if}\:\mathrm{the}\:\mathrm{straight} \\ $$$$\mathrm{lines}\:\bar {\alpha}{z}\:+\:\alpha\bar {{z}}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{and}\:\bar {\beta}{z}\:+\:\beta\bar {{z}}\:+\:{k}\:=\:\mathrm{0} \\ $$$$\mathrm{are}\:\mathrm{parallel}\:\mathrm{and}\:\mathrm{perpendicular}\:\mathrm{according} \\ $$$$\mathrm{as}\:\bar {\alpha}\beta\:−\:\alpha\bar {\beta}\:=\:\mathrm{0}\:\mathrm{and}\:\bar {\alpha}\beta\:+\:\alpha\bar {\beta}\:=\:\mathrm{0}. \\ $$
Answered by ajfour last updated on 12/Aug/17
complex slope μ_1 =−(α/α^� )   μ_2 =−(β/β^� )   so  μ_1 =μ_2    ⇒   αβ^� −α^� β =0  and   μ_1 +μ_2 =0  ⇒   αβ^� +α^� β =0 .
$$\mathrm{complex}\:\mathrm{slope}\:\mu_{\mathrm{1}} =−\frac{\alpha}{\bar {\alpha}} \\ $$$$\:\mu_{\mathrm{2}} =−\frac{\beta}{\bar {\beta}}\: \\ $$$$\mathrm{so}\:\:\mu_{\mathrm{1}} =\mu_{\mathrm{2}} \:\:\:\Rightarrow\:\:\:\alpha\bar {\beta}−\bar {\alpha}\beta\:=\mathrm{0} \\ $$$$\mathrm{and}\:\:\:\mu_{\mathrm{1}} +\mu_{\mathrm{2}} =\mathrm{0}\:\:\Rightarrow\:\:\:\alpha\bar {\beta}+\bar {\alpha}\beta\:=\mathrm{0}\:. \\ $$
Commented by Tinkutara last updated on 12/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 12/Aug/17
it is a matter of complex slopes..
$$\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{matter}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{slopes}.. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *